*p*divides

*x*-

^{p}*x*where

*x*is a natural number (Fermat's "little" theorem). If

*x*and

*y*are relatively prime natural numbers, then

*x*

^{φ(y)}≡1(mod

*y*) where φ(

*y*) is Euler's "phi" function. The greatest common divisor of natural numbers

*x*and

*y*will be denoted by gcd(

*x*,

*y*). A more succinct way of saying that a natural number

*x*is congruent to the

*p*th power of a natural number modulo a natural number

*y*is to say that

*x*is a

*p*th power modulo

*y*.

#### Variants of Fermat's Last Theorem

Let*a*,

*b*, and

*c*be natural numbers relatively prime in pairs and let

*p*be an odd prime. Every prime factor of (

*a*+

^{p}*b*)/(

^{p}*a*+

*b*) other than

*p*is of the form

*pk*+1.

*p*(and no higher power of

*p*) divides (

*a*)/(

^{p}+b^{p}*a+b*) if and only if

*p*divides

*a+b*. Let

*q*be a natural number.

*q*will be said to be a

*p*th power with respect to (

*a*)/(

^{p}+b^{p}*a+b*) if

*q*

^{(f-1)/p}≡1(mod

*f*) for every prime factor

*f*,

*f*≠

*p*, of (

*a*)/(

^{p}+b^{p}*a+b*). Let [(

*a*)/(

^{p}+b^{p}*a+b*)] denote (

*a*)/(

^{p}+b^{p}*a+b*)/

*p*if

*p*divides

*a+b*, or (

*a*)/(

^{p}+b^{p}*a+b*) otherwise. Similarly, let [

*a+b*] denote (

*a+b*)/

*p*if

*p*divides

*a+b*, or

*a+b*otherwise. The following two conjectures are the main topic of this article;

(1) If

*p*>3, there do not exist

*a*and

*b*such that [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power.

(2) If

*p*>3, there do not exist

*a*and

*b*such that [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power, 2

*p*does not divide

*a*,

*b*,

*a-b*, or

*a+b*, and

*a*,

*b*,

*a-b*, or [

*a+b*] is a

*p*th power.

If the first conjecture is true, there are no solutions of Fermat's equation

*a*=

^{p}+b^{p}*c*(which, of course, is already known). The second conjecture encompasses the first case of Fermat's Last Theorem (where

^{p}*p*does not divide

*abc*). (In 1810, Barlow proved that

*a*=

^{p}+b^{p}*c*only if [(

^{p}*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power.) Let

*T*be a natural number. Since

*a*,

*b*, and

*T*are not symmetrical in the equation [(

*a*)/(

^{p}+b^{p}*a+b*)]=

*T*, it is not obvious how to apply the theory of elliptic curves to these problems. The "

^{p}*p*th power w.r.t." concept and the identity

*a*+

^{p}*b*=(

^{p}*a*)+2

^{p}-b^{p}*b*play a central role in proving these conjectures. For example, if

^{p}*p*divides

*a+b*and (

*a*)/(

^{p}+b^{p}*a+b*)/

*p*is a

*p*th power, then

*p*(

*a+b*)/2 is a

*p*th power w.r.t. (

*a*)/(

^{p}-b^{p}*a*-

*b*) (this is a succinct way of saying that (

*p*(

*a+b*))

^{(f-1)/p}≡2

^{(f-1)/p}(mod

*f*) for every prime factor

*f*,

*f*≠

*p*, of (

*a*)/(

^{p}-b^{p}*a-b*)). If

*p*is not a

*p*th power modulo a prime

*f*of the form

*pk*+1, then, for example, if

*f*does not divide

*a+b*, (

*p*(

^{x}*a+b*))

^{(f-1)/p}≡1(mod

*f*), 0≤

*x*<

*p*, has a solution (

*x*defines a congruence class). Furthermore, if 2 is not a

*p*th power modulo

*f*, then, for example, (2

*(*

^{y}*a+b*))

^{(f-1)/p}≡1(mod

*f*), 0≤

*y*<

*p*, has a solution and

*a+b*can be eliminated from the congruences. The objective in the following is to eliminate

*a*,

*b*,

*a-b*, and

*a+b*from certain congruences so that congruence relationships involving only 2 and

*p*are obtained. When

*p*=3, there do exist

*a*and

*b*such that [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power and many properties of such

*a*and

*b*, among them reformulated versions of the classical Furtwängler and Vandiver theorems for Fermat's equation, can be empirically derived. In the following, these "propositions" are stated as if they were true for all

*p*. One justification for doing this is the first conjecture above. Also, more properties of hypothetical solutions of Fermat's equation are shared by solutions of the equation [(

*a*)/(

^{p}+b^{p}*a+b*)]=

*T*,

^{p}*p*=3. For example, it can be easily proved that

*a*=

^{p}+b^{p}*c*implies 2 is a

^{p}*p*th power w.r.t. (

*a*)/(

^{p}-b^{p}*a-b*). Based on empirical evidence collected for

*p*=3, if [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power and 2

*p*divides

*a-b*or

*a+b*, then 2 is a

*p*th power w.r.t. (

*a*)/(

^{p}-b^{p}*a-b*) (although there is no apparent reason why this should be true).

#### Congruence Properties of Prime
Factors of [(*a*^{p}-b^{p})/(*a-b*)]
when [(*a*^{p}+b^{p})/(*a+b*)]
is a *p*th Power

The following propositions are based
on data collected for ^{p}-b

^{p}

^{p}+b

^{p}

*p*=3;

(3) If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power and 2

*p*does not divide

*a*,

*b*,

*a-b*, or

*a+b*, then 2 and

*p*are not

*p*th powers w.r.t. (

*a*)/(

^{p}-b^{p}*a-b*).

(4) If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power, 2

*p*does not divide

*a*,

*b*,

*a-b*, or

*a+b*, and

*f*is a prime factor of [(

*a*)/(

^{p}-b^{p}*a-b*)], then 2 is a

*p*th power modulo

*f*if and only if

*f*is of the form

*p*

^{2}

*k*+1.

By these two propositions, if [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power and 2

*p*does not divide

*a*,

*b*,

*a-b*, or

*a+b*, then there is at least one prime factor of [(

*a*)/(

^{p}-b^{p}*a-b*)] not of the form

*p*

^{2}

*k*+1. In 1912, Furtwängler proved that if

*a*=

^{p}+b^{p}*c*,

^{p}*q*divides

*a*and

*p*does not divide

*ac*, or

*q*divides

*b*and

*p*does not divide

*bc*, then

*q*

^{p}^{-1}≡1(mod

*p*

^{2}). (Proofs of this theorem use the condition that

*a+b*must be a

*p*th power and it is not obvious how to prove a reformulated version of the theorem without using this condition.) Furtwängler also proved that if

*a*=

^{p}+b^{p}*c*,

^{p}*q*divides

*a-b*or

*a+b*, and

*p*does not divide

*a-b*or

*a+b*, then

*q*

^{p}^{-1}≡1(mod

*p*

^{2}). Note that if

*p*does not divide a natural number

*d*, then

*d*

^{p}^{(p-1)}≡1(mod

*p*

^{2}) by Euler's theorem. Then if

*q*

^{p}^{-1}≡1(mod

*p*

^{2}),

*q*is a

*p*th power modulo

*p*

^{2}. The reformulated version of Furtwängler's theorems is;

(5) If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power and 2 does not divide

*a*, then

*p*does not divide

*a*and every prime factor of

*a*is a

*p*th power modulo

*p*

^{2}. If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power, 2 divides

*a*, and

*p*does not divide

*a*, then

*a*/2 is a

*p*th power modulo

*p*

^{2}. Analogous results hold for

*b*. If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power and 2 does not divide

*a-b*, then

*p*does not divide

*a-b*and every prime factor of

*a-b*is a

*p*th power modulo

*p*

^{2}. If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power and 2 does not divide

*a+b*, then

*p*

^{2}does not divide

*a+b*and every prime factor of

*a+b*other than

*p*(if

*p*divides

*a+b*) is a

*p*th power modulo

*p*

^{2}.

The peculiar form of Furtwängler's second theorem, that is, the condition that

*p*not divide

*a-b*or

*a+b*, makes sense when viewed from this perspective; 2 divides

*a-b*if and only if 2 divides

*a+b*. This proposition implies that if [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power,

*p*divides

*a*,

*b*,

*a-b*, or

*a+b*, and 2

*p*does not divide

*a*,

*b*,

*a-b*, or

*a+b*, then 2 divides

*a*or

*b*, and

*p*divides

*a+b*. Note that this proposition implies "split" 2 and

*p*are not possible when

*a*=

^{p}+b^{p}*c*,

^{p}*p*divides

*c*, since

*p*

^{2}cannot divide

*a+b*in this case. (By Barlow's formulas,

*p*(

*a+b*) must be a

*p*th power when

*a*=

^{p}+b^{p}*c*,

^{p}*p*divides

*c*.) The requirement that 2

*p*divide

*a+b*could be said to be a characteristic property of the equation

*a*=

^{p}+b^{p}*c*,

^{p}*p*divides

*c*. A related proposition is;

(6) If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power, then

*p*

^{2}divides

*a*if 2

*p*divides

*a*,

*p*

^{2}divides

*b*if 2

*p*divides

*b*,

*p*

^{2}divides

*a-b*if 2

*p*divides

*a-b*, or

*p*

^{3}divides

*a+b*if 2

*p*divides

*a+b*.

In the following,

*p*/2 will be said to be a

*p*th power modulo a prime

*f*if

*p*

^{(f-1)/p}≡ 2

^{(f-1)/p}(mod

*f*). More propositions are;

(7) If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power, 2 divides

*a*or

*b*, and

*f*is a prime factor of [(

*a*)/(

^{p}-b^{p}*a-b*)] not of the form

*p*

^{2}

*k*+1, then exactly one of 2

*p*,

*p*, or

*p*/2 is a

*p*th power modulo

*f*.

(8) If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power,

*p*divides

*a+b*, and

*f*is a prime factor of [(

*a*)/(

^{p}-b^{p}*a-b*)] of the form

*p*

^{2}

*k*+1, then

*pa*,

*pb*,

*p*

^{2}(

*a-b*), and

*p*(

*a+b*) are

*p*th powers modulo

*f*. If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power,

*p*divides

*a*,

*b*, or

*a-b*, and

*f*is a prime factor of [(

*a*)/(

^{p}-b^{p}*a-b*)] of the form

*p*

^{2}

*k*+1, then

*a*,

*b*,

*p*(

*a-b*), and

*a+b*are

*p*th powers modulo

*f*.

Note that

*p*is not precluded from being a

*p*th power modulo

*f*in Proposition (8). If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power and 2

*p*does not divide

*a*,

*b*,

*a-b*, or

*a+b*, there is apparently nothing to prevent

*p*from being a

*p*th power modulo every prime factor of [(

*a*)/(

^{p}-b^{p}*a-b*)] not of the form

*p*

^{2}

*k*+1. By these propositions, if [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power, 2

*p*does not divide

*a*,

*b*,

*a-b*, or

*a+b*, and

*p*is a

*p*th power modulo every prime factor of [(

*a*)/(

^{p}-b^{p}*a-b*)] not of the form

*p*

^{2}

*k*+1, then

*a*,

*b*,

*a-b*, and [

*a+b*] are not

*p*th powers. (By Proposition (3), there would be at least one prime factor

*f*of [(

*a*)/(

^{p}-b^{p}*a-b*)] of the form

*p*

^{2}

*k*+1 such that

*p*was not a

*p*th power modulo

*f*. Then by Proposition (8),

*a*,

*b*,

*a-b*, or [

*a+b*] couldn't be a pth power.) More propositions are;

(9) If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power, 2 divides

*a*,

*p*does not divide

*a*,

*f*is a prime factor of [(

*a*)/(

^{p}-b^{p}*a-b*)] not of the form

*p*

^{2}

*k*+1, and

*p*/2 is a

*p*th power modulo

*f*,

*then*

*pa*,

*p*

^{2}

*b*,

*p*(

*a-b*), and

*a+b*are

*p*th powers modulo

*f*. If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power, 2 divides

*a*,

*p*does not divide

*a*,

*f*is a prime factor of [(

*a*)/(

^{p}-b^{p}*a-b*)] not of the form

*p*

^{2}

*k*+1, and 2

*p*is a

*p*th power modulo

*f*, then

*pa*,

*b*,

*a-b*, and

*p*

^{2}(

*a+b*) are

*p*th powers modulo

*f*. Analogous results hold for

*b*.

(10) If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power, 2 divides

*a*, and

*p*does not divide

*a*, then

*pa*, 2

*pb*, 2

^{2}

*p*

^{2}(

*a-b*), and 2

^{2}

*p*(

*a+b*) are

*p*th powers w.r.t. (

*a*)/(

^{p}-b^{p}*a-b*). Also, either 2

*p*is a

*p*th power w.r.t. (

*a*)/(

^{p}-b^{p}*a-b*) or none of 2

*p*,

*p*,

*p*/2, or 2 is a

*p*th power w.r.t. (

*a*)/(

^{p}-b^{p}*a-b*) (if [(

*a*)/(

^{p}-b^{p}*a-b*)] has only one distinct prime factor, then [(

*a*)/(

^{p}-b^{p}*a-b*)] is prime and 2

*p*is a

*p*th power w.r.t. (

*a*)/(

^{p}-b^{p}*a-b*)). Analogous results hold for

*b*.

Note that if 2

*p*is a

*p*th power w.r.t. (

*a*)/(

^{p}-b^{p}*a-b*) in Proposition (10), then

*a*/2,

*b*,

*a-b*, and [

*a+b*] are

*p*th powers w.r.t. (

*a*)/(

^{p}-b^{p}*a-b*). Since Propositions (8), (9), and 10 are based solely on data collected for

*p*=3, their form is sometimes ambiguous in that the

*p*

^{2}and 2

^{2}factors might be

*p*

^{p}^{-1 }and 2

^{p-1}instead. If

*p*(

*a+b*)/2 is a

*p*th

*power w.r.t. [(*

*a*)/(

^{p}-b^{p}*a-b*)] (as implied by a second-case solution of Fermat's equation,

*p*divides

*c*,

*p≥*3),

*f*is a prime factor of [(

*a*)/(

^{p}-b^{p}*a-b*)], and 2

*p*is a

*p*th power modulo

*f*, then

*p*

^{2}(

*a+b*) is a

*p*th power modulo

*f*, so the

*p*

^{2}factor of

*a+b*in Proposition (9) is unambiguous. Propositions (9) and (10) are consistent when

*p*is not a

*p*th power modulo

*f*only if 8 is a

*p*th power modulo

*f*, but 8 can be a

*p*th power modulo

*f*only if

*p*=3 (2

^{f-1}≡1(mod

*f*) and (2

^{3})

^{(f-1)/p}≡1(mod

*f*),

*p*≠3, implies 2

^{(f-1)/p}≡1(mod

*f*) [since in this case, the greatest common divisor of

*f*-1 and 3(

*f*-1)/

*p*is (

*f*-1)/

*p*], a contradiction). This follows from eliminating

*b*,

*a-b*, and

*a+b*from the congruences if 2 divides

*a*, or eliminating

*a*,

*a-b*, and

*a+b*from the congruences if 2 divides

*b*. If a 2

^{p-1}

*p*factor of the

*a+b*term in Proposition (10) had been used, some inconsistency for

*p*>3 could have been avoided (this implies

*p*/2, 2

^{2}

*p*, and 2

^{p-1}

*p*are

*p*th powers modulo

*f*if

*p*/2 is a

*p*th power modulo

*f*, or 2

*p*, 2

^{2}

*p*

^{2}, and 2

^{p-1}/

*p*are

*p*th powers modulo

*f*if 2

*p*is a

*p*th power modulo

*f*). However, using a 2

^{ p-1}

*p*

^{ p-1}, 2

^{2}

*p*

^{p}^{-1}, 2

^{p-1}

*p*

^{2 }, or 2

^{2}

*p*

^{2}factor of the

*a-b*term in Proposition (10) implies that 8 is a

*p*th power modulo

*f*, a contradiction for

*p*>3. This may just mean that one (or both) of the propositions is specific to

*p*=3. It's plausible that the maximum

*p*exponent used in Proposition (9) is related to the number of the terms

*a*,

*b*,

*a-b*, and

*a+b*(and not to the

*p*value itself). If

*p*is not a

*p*th power modulo

*f*, then, for example, (

*p*(

^{x}*a-b*))

^{(f-1)/p}≡1(mod

*f*), 0

*≤x<p*, has a solution, so Proposition (9) should remain the same for

*p*>3. If it's granted that 2

*p*should be a

*p*th power modulo

*f*some of the time (note that this implies that

*p*is not a

*p*th power modulo

*f*), then Proposition (10) should remain the same for

*p*>3. (In this case, eliminating

*b*,

*a-b*, and

*a+b*from the congruences if 2

*divides*

*a*, or eliminating

*a*,

*a-b*, and

*a+b*from the congruences if 2 divides

*b*, gives 2

*p*, 2

^{2}

*p*

^{2}, and

*p*/2

^{2}are

*p*th powers modulo

*f*. Note that 2

^{2}

*p*

^{2}is the square of 2

*p*;

*x*=2 is the only possible solution of [2

^{x}p^{2}(

*a-b*)]

^{(f-1)/p}≡1(mod

*f*) when

*x*=1 is the solution of (2

*)*

^{x}pa^{(f-1)/p}≡1(mod

*f*), 2 divides

*b*, or (2

*)*

^{x}pb^{(f-1)/p}≡1(mod

*f*), 2 divides

*a*.)

*Other propositions are;*

(11) If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power, 2 divides

*a*,

*f*is a prime factor of [(

*a*)/(

^{p}-b^{p}*a-b*)] not of the form

*p*

^{2}

*k*+1, and

*p*is a

*p*th power modulo

*f*, then

*a*, 2

*b*, 2

^{2}(

*a-b*), and 2

^{2}(

*a+b*) (and not 2) are

*p*th powers modulo

*f*. Analogous results hold for

*b*. If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power, 2 divides

*a-b*or

*a+b*,

*f*is a prime factor of [(

*a*)/(

^{p}-b^{p}*a-b*)] not of the form

*p*

^{2}

*k*+1, and

*p*is a

*p*th power modulo

*f*,

*then*

*a-b*,

*a+b*, and 2 (and not

*a*or

*b*) are

*p*th powers modulo

*f*.

(12) If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power, then

*a*is a

*p*th power w.r.t. (

*a*)/(

^{p}-b^{p}*a-b*) if 2

*p*divides

*a*, or

*b*is a

*p*th power w.r.t. (

*a*)/(

^{p}-b^{p}*a-b*) if 2

*p*divides

*b*, or

*p*(

*a-b*) and

*a+b*are

*p*th powers w.r.t. (

*a*)/(

^{p}-b^{p}*a-b*) if 2

*p*divides

*a-b*, or

*p*

^{2}(

*a-b*) and

*p*(

*a+b*) are

*p*th powers w.r.t. (

*a*)/(

^{p}-b^{p}*a-b*) if 2

*p*divides

*a+b*.

(13) If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power and 2

*p*divides

*a-b*or

*a+b*, then 2 is a

*p*th power w.r.t. (

*a*)/(

^{p}-b^{p}*a-b*).

(14) If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power, 2

*p*divides

*a*,

*f*is a prime factor of [(

*a*)/(

^{p}-b^{p}*a-b*)] not of the form

*p*

^{2}

*k*+1, and

*p*/2 is a

*p*th power modulo

*f*, then

*a*,

*pb*,

*a-b*, and

*p*

^{2}(

*a+b*) are

*p*th powers modulo

*f*. If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power, 2

*p*divides

*a*,

*f*is a prime factor of [(

*a*)/(

^{p}-b^{p}*a-b*)] not of the form

*p*

^{2}

*k*+1, and 2

*p*is a

*p*th power modulo

*f*, then

*a*,

*p*

^{2}

*b*,

*p*

^{2}(

*a-b*), and

*p*(

*a+b*) are

*p*th powers modulo

*f*. Analogous results hold for

*b*.

*If [(*

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power, 2

*p*divides

*a-b*,

*f*is a prime factor of [(

*a*)/(

^{p}-b^{p}*a-b*)] not of the form

*p*

^{2}

*k*+1, and

*p*is not a

*p*th power modulo

*f*,

*then (1)*

*pa*,

*p*

^{2}

*b*,

*p*(

*a-b*), and

*a+b*are

*p*th powers modulo

*f*, or (2)

*p*

^{2}

*a*,

*pb*,

*p*(

*a-b*), and

*a+b*are

*p*th powers modulo

*f*. If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power, 2

*p*divides

*a+b*,

*f*is a prime factor of [(

*a*)/(

^{p}-b^{p}*a-b*)] not of the form

*p*

^{2}

*k*+1, and

*p*is not a

*p*th power modulo

*f*,

*then (1)*

*a*,

*p*

^{2}

*b*,

*p*

^{2}(

*a-b*), and

*p*(

*a+b*) are

*p*th powers modulo

*f*,

*or (2)*

*p*

^{2}

*a*,

*b*,

*p*

^{2}(

*a-b*), and

*p*(

*a+b*) are

*p*th powers modulo

*f*.

(15) If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power, 2

*p*divides

*a*or

*b*, [(

*a*)/(

^{p}-b^{p}*a-b*)] has two distinct prime factors, and neither distinct prime factor is of the form

*p*

^{2}

*k*+1, then [(

*a*)/(

^{p}-b^{p}*a-b*)] is of the form

*p*

^{2}

*k*+1, and (1) 2

*p*is a

*p*th power modulo both distinct prime factors, or (2)

*p*is a

*p*th power modulo both distinct prime factors, or (3)

*p*/2 is a

*p*th power modulo both distinct prime factors. If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power, 2

*p*divides

*a-b*or

*a+b*, [(

*a*)/(

^{p}-b^{p}*a-b*)] has two distinct prime factors, and neither distinct prime factor is of the form

*p*

^{2}

*k*+1, then [(

*a*)/(

^{p}-b^{p}*a-b*)] is of the form

*p*

^{2}

*k*+1.

In this section, it is assumed that [(

*a*-

^{p}*b*)/(

^{p}*a-b*)] can't be a

*p*th power when [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power. (As will be shown, the congruence properties of the prime factors of [(

*a*)/(

^{p}+b^{p}*a+b*)] when [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power are similar to the congruence properties of the prime factors of [(

*a*)/(

^{p}-b^{p}*a-b*)] when [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power, but are not the same.)

Let

*T*be a natural number. If

*p*=3, every prime factor of

*T*is of the form

*pk*+1, and

*T*has

*n*such distinct prime factors, then

*T*or

^{p}*pT*has exactly

^{p}*pn*representations of the form (

*a*)/(

^{p}+b^{p}*a+b*). Proving the first conjecture when

*p*divides

*a+b*would entail proving that if one representation of

*pT*of the form (

^{p }*a*)/(

^{p}+b^{p}*a+b*) exists, then other representations exist and that 2 and

*p*split for some of these representations. There is little evidence that there would exist different representations of

*pT*of the form (

^{p}*a*)/(

^{p}+b^{p}*a+b*) for

*p*>3. Even if there were a representation ((

*a*')

^{p}+(

*b*')

^{p})/(

*a*'+

*b*') with split 2 and

*p*, how to deal with the case where

*p*was a

*p*th power modulo every prime factor of [((

*a*')

^{p}-(

*b*')

^{p})/(

*a*'-

*b*')] not of the form

*p*

^{2}

*k*+1 is unknown.

#### Congruence Properties of
Prime [(*a*^{p}-b^{p})/(*a-b*)]
when [(*a*^{p}+b^{p})/(*a*+*b*)]
is a *p*th Power

In this section, more empirical
evidence in support of Propositions
(8), (12), and (13) is given. The
following propositions are based on
data collected for ^{p}-b

^{p}

^{p}+b

^{p}

*p*=3, 5, 7, and 11;

(16) If

*p*>3,

*p*

^{2}divides

*a*,

*b*,

*a-b*, or

*a+b*, [(

*a*)/(

^{p}+b^{p}*a+b*)] and [(

*a*)/(

^{p}-b^{p}*a-b*)] are primes of the form

*p*

^{2}

*k*+1, and 2 is a

*p*th power modulo [(

*a*)/(

^{p}+b^{p}*a+b*)] or [(

*a*)/(

^{p}-b^{p}*a-b*)], then [(

*a*)/(

^{p}-b^{p}*a-b*)] is a

*p*th power modulo [(

*a*)/(

^{p}+b^{p}*a+b*)] and [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power modulo [(

*a*)/(

^{p}-b^{p}*a-b*)].

*An analogous result holds for*

*p*=3 if

*p*

^{2}divides

*a*or

*b*or

*p*

^{3}divides

*a-b*or

*a+b*.

(17) If [(

*a*)/(

^{p}-b^{p}*a-b*)] is prime,

*q*

^{p}^{-1}≡1(mod

*p*

^{2}), and

*q*divides

*a*,

*b*,

*a+b*, or

*a-b*, then

*q*is a

*p*th power modulo [(

*a*)/(

^{p}-b^{p}*a-b*)].

(18) If

*p*>3, [(

*a*)/(

^{p}-b^{p}*a-b*)] is prime, and

*p*

^{2}divides

*a*,

*b*,

*a+b*, or

*a-b*, then

*p*is a

*p*th power modulo [(

*a*)/(

^{p}-b^{p}*a-b*)]. If

*p*=3, [(

*a*)/(

^{p}-b^{p}*a-b*)] is prime, and

*p*

^{2}divides

*a*,

*b*, or

*a+b*or

*p*

^{3}divides

*a-b*, then

*p*is a

*p*th power modulo [(

*a*)/(

^{p}-b^{p}*a-b*)].

(19) If [(

*a*)/(

^{p}-b^{p}*a-b*)] is prime,

*p*does not divide

*q*,

*p*divides

*a*and

*q*divides

*a*, or

*p*divides

*b*and

*q*divides

*b*, or

*p*divides

*a+b*and

*q*divides

*a+b*or

*a-b*, then

*q*is a

*p*th power modulo [(

*a*)/(

^{p}-b^{p}*a-b*)]. If

*p*>3, [(

*a*)/(

^{p}-b^{p}*a-b*)] is prime,

*p*does not divide

*q*,

*p*divides

*a-b*, and

*q*divides

*a+b*or

*a-b*, then

*q*is a

*p*th power modulo [(

*a*)/(

^{p}-b^{p}*a-b*)]. If

*p*=3, [(

*a*)/(

^{p}-b^{p}*a-b*)] is prime,

*p*does not divide

*q*,

*p*

^{2}divides

*a-b*, and

*q*divides

*a+b*or

*a-b*, then

*q*is a

*p*th power modulo [(

*a*)/(

^{p}-b^{p}*a-b*)].

(20) If

*p*>3, [(

*a*)/(

^{p}-b^{p}*a-b*)] is a prime of the form

*p*

^{2}

*k*+1, and

*p*

^{2}divides

*a*,

*b*,

*a+b*, or

*a-b*, then

*a*,

*b*,

*a+b*,

*a-b*, and

*p*are

*p*th powers modulo [(

*a*)/(

^{p}-b^{p}*a-b*)]. If

*p*=3, [(

*a*)/(

^{p}-b^{p}*a-b*)] is a prime of the form

*p*

^{2}

*k*+1, and

*p*

^{2}divides

*a*,

*b*, or

*a+b*or

*p*

^{3}divides

*a-b*, then

*a*,

*b*,

*a+b*,

*a-b*, and

*p*are

*p*th powers modulo [(

*a*)/(

^{p}-b^{p}*a-b*)].

Propositions (17), (18), and (19) and Propositions (5) and (6) lead to the following proposition;

(21) If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power, 2

*p*divides

*a*,

*b*,

*a-b*, or

*a+b*, and [(

*a*)/(

^{p}-b^{p}*a-b*)] is prime, then every factor of

*a*,

*b*,

*a-b*, and

*a+b*is a

*p*th power modulo [(

*a*)/(

^{p}-b^{p}*a-b*)] (note that this implies [(

*a*)/(

^{p}-b^{p}*a-b*)] is a prime of the form

*p*

^{2}

*k*+1).

Based on data collected for

*p*=3, generalized versions of Propositions (17), (18), and (19) are true when [(

*a*)/(

^{p}-b^{p}*a-b*)]=

*U*where

^{k}*U*is a prime and

*p*does not divide

*k*. (The propositions would be modified so that the modulus would be

*U*instead of [(

*a*)/(

^{p}-b^{p}*a-b*)].) This gives the following proposition;

(22) If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power, 2

*p*divides

*a*,

*b*,

*a-b*, or

*a+b*, and [(

*a*)/(

^{p}-b^{p}*a-b*)]=

*U*where

^{k}*U*is a prime and

*p*does not divide

*k*, then every factor of

*a*,

*b*,

*a-b*, and

*a+b*is a

*p*th power modulo

*U*.

#### Wieferich's Criterion and the
"*p*th Power with Respect to" Concept

In 1909, Wieferich proved that if*a*=

^{p}+b^{p}*c*,

^{p}*p*does not divide

*abc*, then 2

^{ p-1}≡1(mod

*p*

^{2}). Wieferich derived this criterion from very complicated formulas; a simpler approach is to employ the "

*p*th power w.r.t." concept. The following proposition is based on data collected for

*p*=3, 5, 7, and 11;

(23) If

*p*>3,

*q*

^{p}^{-1}≠1(mod

*p*

^{2}), and

*q*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*), then

*p*divides

*a*if

*q*divides

*a*,

*p*divides

*b*if

*q*divides

*b*, or

*p*divides

*a-b*or

*a+b*if

*q*divides

*a-b*or

*a+b*. If

*p*=3,

*q*

^{p}^{-1}≠1(mod

*p*

^{2}), and

*q*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*), then

*p*divides

*a-b*or

*a+b*if

*q*divides

*a-b*or

*a+b*.

This proposition precludes first-case solutions of Fermat's equation except when 2

^{p-1}≡1(mod

*p*

^{2}) or

*p*=3 since

*a*=

^{p}+b^{p}*c*implies

^{p}*c*divides

^{p}+b^{p}*a*2

^{p}+*b*,

^{p}*c*divides 2

^{p}+a^{p}*a*, and

^{p}+b^{p}*a*divides

^{p}-b^{p}*c*-2

^{p}*a*and hence that 2 is a

^{p}*p*th power w.r.t. (

*c*)/(

^{p}+b^{p}*c+b*), (

*c*)/(

^{p}+a^{p}*c+a*), and (

*a*)/(

^{p}-b^{p}*a-b*). (If

*a*=

^{p}+b^{p}*c*, one of

^{p}*a*,

*b*, and

*c*must be even.) Mirimanoff proved that a first-case solution of Fermat's equation implies that 3

^{p-1}≡1(mod

*p*

^{2}). The probability that a root of the congruence

*x*

^{p}^{-1}≡1(mod

*p*

^{2}), 0<

*x*<

*p*

^{2}, is one larger than another root is 1/

*p*(since there are

*p*-1 roots having

*p*(

*p*-1) possible values). There then shouldn't be any

*p*such that 3

^{p-1}≡2

^{p-1}≡1(mod

*p*

^{2}) since the sum of (1/

*p*)(1/

*p*) over all

*p*converges and the only

*p*less than 3x10

^{9}such that 2

^{p-1}≡1(mod

*p*

^{2}) are 1093 and 3511, and 3

^{p-1}≠1(mod

*p*

^{2}) for either of these

*p*.

Let ζ be a primitive

*p*th root of unity and

*K*=

*Q*(ζ), a cyclotomic field of degree

*p*-1 over

*Q*. Let λ denote 1-ζ. The following proposition follows from the Chinese remainder theorem (and has also been confirmed using data collected for

*p*=3);

(24)

*q*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*) if and only if

*q*is congruent to the

*p*th power of an integer modulo

*a*+

*ζb*,

*q*is congruent to the

*p*th power of an integer modulo

*a*+ζ

^{2}

*b*,

*q*is congruent to the

*p*th power of an integer modulo

*a*+ζ

^{3}

*b*, ..., and

*q*is congruent to the

*p*th power of an integer modulo

*a*+ζ

^{p-1}

*b*(these are integers in

*K*).

####
Barlow's Formulas and the "*p*th Power
with Respect to" Concept

Barlow's formulas implied by a
first-case solution of Fermat's
equation are (*c*)/(

^{p}-b^{p}*c-b*)=

*R*, (

^{p}*c*)/(

^{p}-a^{p}*c-a*)=

*S*, (

^{p}*a*)/(

^{p}+b^{p}*a+b*)=

*T*,

^{p}*c-b*=

*r*,

^{p}*c-a*=

*s*, and

^{p}*a+b*=

*t*where

^{p}*rR*=

*a*,

*sS*=

*b*,

*tT*=

*c*, and gcd(

*r*,

*R*)=gcd(

*s*,

*S*)=gcd(

*t*,

*T*)=1. Then

*a*divides

*S*

^{p}-c^{p}^{-1}and

*T*

^{p}-b^{p}^{-1}and hence

*c*

^{(f-1)/p}≡1(mod

*f*) and

*b*

^{(f-1)/p}≡1(mod

*f*) for every prime factor

*f*of

*R*. (Note that

*c*

^{(f-1)/p}≡

*b*

^{(f-1)/p}(mod

*f*) and

*c*≡

^{p}*b*(mod

^{p}*f*) so that every prime factor of (

*c*)/(

^{p}-b^{p}*c-b*) must be of the form

*p*

^{2}

*k*+1 [first proved by Sophie Germain].) Analogous results hold for

*b*and

*c*. There can be first-case solutions of Fermat's equation only if

*c*,

*b*, and

*c-b*are

*p*th powers w.r.t. (

*c*)/(

^{p}-b^{p}*c-b*),

*c*,

*a*, and

*c-a*are

*p*th powers w.r.t. (

*c*)/(

^{p}-a^{p}*c-a*),

*a*,

*b*, and

*a+b*are

*p*th powers w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*), and every prime factor of (

*c*)/(

^{p}-b^{p}*c-b*), (

*c*)/(

^{p}-a^{p}*c-a*), and (

*a*)/(

^{p}+b^{p}*a+b*) is of the form

*p*

^{2}

*k*+1. Also,

*r*divides

*s*,

^{p}-t^{p}*s*divides

*r*, and

^{p}-t^{p}*t*divides

*r*. If

^{p}+s^{p}*n*>2, [(

*x*)

^{n}+y^{n}^{1/n}-

*x*]

^{1/n}+[(

*x*)

^{n}+y^{n}^{1/n}-

*y*]

^{1/n}>(

*x+y*)

^{1/n}where

*x*and

*y*are positive real numbers, therefore

*r+s*>

*t*>

*r*,

*s*and hence each of

*r*,

*s*, and

*t*has a prime factor of the form

*pk*+1. Also, (

*a*)/(

^{p}+b^{p}*a+b*)=(

*a+b*)(

*a*

^{p}^{-2}-2

*a*

^{p}^{-3}b+3

*a*

^{p}^{-4}

*b*

^{2}-...-(

*p*-1)

*b*

^{p}^{-2})+

*pb*

^{p}^{-1}so that

*pb*

^{p}^{-1}≡

*T*(mod

^{p}*a+b*).

Barlow's formulas implied by a second-case solution of Fermat's equation where

*p*divides

*c*are (

*c*)/(

^{p}-b^{p}*c-b*)=

*R*, (

^{p}*c*)/(

^{p}-a^{p}*c-a*)=

*S*, (

^{p}*a*)/(

^{p}+b^{p}*a+b*)=

*pT*,

^{p}*c-b*=

*r*,

^{p}*c-a*=

*s*, and

^{p}*a+b*=(

*p*)

^{k}t^{p}/

*p*where

*rR*=

*a*,

*sS*=

*b*,

*p*=

^{k}tT*c*, and gcd(

*r*,

*R*)=gcd(

*s*,

*S*)=gcd(

*p*,

^{k}t*T*)=1. Then

*c*divides

*R*-

^{p}*b*

^{p}^{-1 }and

*S*

^{p}-a^{p}^{-1 }and hence

*b*

^{(f-1)/p}≡1(mod

*f*) and

*a*

^{(f-1)/p}≡1(mod

*f*) for every prime factor

*f*of

*T*. Also,

*b*divides

*pT*

^{p}-a^{p}^{-1}and

*R*-

^{p}*c*

^{p}^{-1}and hence (

*pa*)

^{(f-1)/p}≡1(mod

*f*) and

*c*

^{(f-1)/p}≡1(mod

*f*) for every prime factor

*f*of

*S*. Analogous results hold for

*a*. There can be second-case solutions of Fermat's equation where

*p*divides

*c*only if

*c*,

*pb*, and

*c-b*are

*p*th powers w.r.t. (

*c*)/(

^{p}-b^{p}*c-b*),

*c*,

*pa*, and

*c-a*are

*p*th powers w.r.t. (

*c*)/(

^{p}-a^{p}*c-a*),

*a*,

*b*, and

*p*(

*a+b*) are

*p*th powers w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*), and every prime factor of (

*a*)/(

^{p}+b^{p}*a+b*)/

*p*is of the form

*p*

^{2}

*k*+1 (based on these formulas, the prime factors of (

*c*)/(

^{p}-b^{p}*c-b*) and (

*c*)/(

^{p}-a^{p}*c-a*) are not necessarily of the form

*p*

^{2}

*k*+1). (The requirement that every prime factor of (

*a*)/(

^{p}+b^{p}*a+b*)/

*p*be of the form

*p*

^{2}

*k*+1 could be said to be another characteristic property of the equation

*a*=

^{p}+b^{p}*c*,

^{p}*p*divides

*c*.) Also,

*p*divides

*a*

^{p}^{-2}-2

*a*

^{p}^{-3}

*b*+3

*a*

^{p}^{-4}

*b*

^{2}-...-(

*p*-1)

*b*

^{p}^{-2}so that

*b*

^{p}^{-1}≡

*T*(mod

^{p}*a+b*) (this is relevant to fractional ideals to be discussed in the next section). The following proposition is based on data collected for

*p*=3, 5, 7, and 11;

(25) If

*p*>3,

*a*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*), and

*p*does not divide

*a*,

*then*

*a*

^{p}^{-1}≡1(mod

*p*

^{2}). If

*p*=3,

*a*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*), and

*p*divides

*b*or

*a-b*or

*p*

^{2}divides

*a+b*, then

*a*

^{p}^{-1}≡1(mod

*p*

^{2}). Analogous results hold for

*b*. If

*a-b*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*) and

*p*does not divide

*a-b*or

*a+b*, then (

*a-b*)

^{p-1}≡1(mod

*p*

^{2}). Analogous results hold for

*a+b*.

Proposition (25) shows that a second-case solution of Fermat's equation where

*p*divides

*c*implies

*a*

^{p}^{-1}≡1(mod

*p*

^{2}) and

*b*

^{p}^{-1}≡1(mod

*p*

^{2}) (avoiding the constraint that

*p*not divide

*c*in Furtwängler's first theorem).

#### Furtwängler's Theorems and Hasse's Reciprocity Law

This section requires some familiarity with algebraic number theory. In the following, the brackets "{}" denote the*p*th power residue symbol. Hasse used one of his reciprocity laws to give a more systematic proof of Furtwängler's theorems. Hasse's reciprocity law is; {β/α}{α/β}

^{-1}=ζ

^{Tr(η)}where η=((β-1)/

*p*)((α-1)/λ) for all α, β in

*Q*(ζ) with gcd(α, β)=1, α≡1(mod λ), β≡1(mod

*p*), and where Tr denotes the trace from

*Q*(ζ) to

*Q*. Setting α to (

*a*+ζ

*b*)/(

*a+b*) and β to

*q*

^{p}^{-1}where

*q*divides

*b*gives {

*q*

^{p}^{-1}/[(

*a*+ζ

*b*)/(

*a+b*)]}=ζ

^{-u(p-1)b/(a+b)}where

*u*=(

*q*

^{p}^{-1}-1)/

*p*since α=1-

*b*λ/(

*a+b*)≡1(mod λ) and α≡1(mod

*q*). If

*a*=

^{p}+b^{p}*c*,

^{p}*p*does not divide

*c*, then α is the

*p*th power of an ideal and hence {β/α}=1 for all β in

*Q*(ζ) that are prime to α. Then if

*p*does not divide

*b*,

*p*must divide

*u*.

The question of which of (

*q*

^{p}^{-1}-1)/

*p*and

*a+b*is divisible by the largest power of

*p*can be avoided by considering the reciprocal of α. ((

*a*+ζ

^{2}

*b*)/(

*a*+ζ

*b*))((

*a*+ζ

^{3}

*b*)/(

*a*+ζ

^{2}

*b*))···((

*a*+

*ζ*)/(

^{p}b*a*+ζ

^{p-1}

*b*))=(

*a+b*)/(

*a*+ζ

*b*). Let

*d*(

*m*,

*n*) denote

*m*"choose"

*n*(a binomial coefficient). Let

*f*

_{1}denote

*d*(1, 0)

*a*

^{p}^{-2}-

*d*(2, 1)

*a*

^{p}^{-3}

*b*+

*d*(3, 2)

*a*

^{p}^{-4}

*b*

^{2}-...-

*d*(

*p*-1,

*p*-2)

*b*

^{p}^{-2},

*f*

_{2}denote

*d*(2, 0)

*a*

^{p}^{-3}-

*d*(3, 1)

*a*

^{p}^{-4}

*b*+

*d*(4, 2)

*a*

^{p}^{-5}

*b*

^{2}-...+

*d*(

*p*-1,

*p*-3)

*b*

^{p}^{-3},

*f*

_{3}denote

*d*(3, 0)

*a*

^{p}^{-4}-

*d*(4, 1)

*a*

^{p}^{-5}

*b*+

*d*(5, 2)

*a*

^{p}^{-6}

*b*

^{2}-...-

*d*(

*p*-1,

*p*-4)

*b*

^{p}^{-4}, ..., and

*f*

_{p}_{-1}denote

*d*(

*p*-1, 0). Collecting terms in the product (1-

*b*λζ/(

*a*+ζ

*b*))(1-

*b*λζ

^{2}/(

*a*+ζ

^{2}

*b*))···(1-

*b*λζ

^{p-1}/(

*a*+ζ

^{p-1}

*b*)) gives (

*a*+

*b*)/(

*a*+ζ

*b*)=1+(

*bλf*

_{1}+

*b*

^{2}λ

^{2}

*f*

_{2}+...+

*b*

^{p}^{-1}λ

^{p-1}

*f*

_{p-1})/((

*a*)/(

^{p}+b^{p}*a+b*)). Tr(1)=

*p*-1 and Tr(λ

^{k})=

*p*. Set α to (

*a+b*)/(

*a*+ζ

*b*). Substituting for Tr(1), Tr(λ), Tr(λ

^{2}), ..., Tr(λ

^{p-2}) and collecting terms gives Tr((α-1)/λ)=

*b*((

*p*-1)

*a*

^{p}^{-2}-(

*p*-2)

*a*

^{p}^{-3}

*b*+(

*p*-3)

*a*

^{p}^{-4}

*b*

^{2}-...-

*b*

^{p}^{-2})/((

*a*)/(

^{p}+b^{p}*a+b*)). Setting β to

*q*

^{p}^{-1}where

*q*divides

*b*gives {

*q*

^{p}^{-1}/[(

*a+b*)/(

*a*+ζ

*b*)]}=ζ

^{uv}where

*v*=

*b*((

*p*-1)

*a*

^{p}^{-2}-(

*p*-2)

*a*

^{p}^{-3}

*b*+(

*p*-3)

*a*

^{p}^{-4}

*b*

^{2}-...-

*b*

^{p}^{-2})/((

*a*)/(

^{p}+b^{p}*a+b*)). If

*p*divides

*a+b*, then

*p*also divides (

*a*)/(

^{p}+b^{p}*a+b*). (

*p*-1)

*a*

^{p}^{-2}-(

*p*-2)

*a*

^{p}^{-3}

*b*+(

*p*-3)

*a*

^{p}^{-4}

*b*

^{2}-...-

*b*

^{p}^{-2}is congruent to -(

*a+b*)

^{p-2}modulo

*p*, therefore if

*p*divides

*a+b*, then 1/

*p*does not divide

*v*. By Proposition (25), if

*a*=

^{p}+b^{p}*c*,

^{p}*p*divides

*c*, then

*b*

^{p}^{-1}≡1(mod

*p*

^{2}) and hence {

*b*

^{p}^{-1}/[(

*a+b*)/(

*a*+ζ

*b*)]}=1.

*p*=(1-ζ)(1-ζ

^{2})(1-ζ

^{3})···(1-ζ

^{p-1}) and the ideals [1-ζ], [1-ζ

^{2}], [1-ζ

^{3}], ..., [1-ζ

^{p-1}] are equal. If

*p*divides

*a+b*and (

*a*)/(

^{p}+b^{p}*a+b*)/

*p*is a

*p*th power, then ((

*a*+ζ

^{2}

*b*)/(1-ζ

^{2}))/((

*a*+ζ

*b*)/(1-ζ)), ((

*a*+ζ

^{3}

*b*)/(1-ζ

^{3}))/((

*a*+ζ

^{2}

*b*)/(1-ζ

^{2})), ((

*a*+ζ

^{4}

*b*)/(1-ζ

^{4})/((

*a*+ζ

^{3}

*b*)/(1-ζ

^{3})), ..., ((

*a*+ζ

^{p-1}

*b*)/(1-ζ

^{p-1}))/((

*a*+ζ

^{p-2}

*b*)/(1-ζ

^{p-2})) are

*p*th powers of fractional ideals and hence (

*a*+ζ

^{2}

*b*)/(

*a*+ζ

*b*), (

*a*+ζ

^{3}

*b*)/(

*a*+ζ

^{2}

*b*), (

*a*+ζ

^{4}

*b*)/(

*a*+ζ

^{3}

*b*), ..., (

*a*+ζ

^{p-1}

*b*)/(

*a*+ζ

^{p-2}

*b*) are

*p*th powers of fractional ideals. Furthermore, (

*a*+ζ

^{i+1}

*b*)/(

*a*+

*ζ*)=1-

^{i}b*b*λζ

^{i}/(

*a*+

*ζ*)=α

^{i}b_{i}≡1(mod λ),

*i*=1, 2, 3, ...,

*p*-2, and hence {β/α

_{i}}=1 for all β in

*Q*(ζ) that are prime to α

_{i}. Tr(-

*b*ζ/(

*a*+ζ

*b*))+Tr(-

*b*ζ

^{2}/(

*a*+ζ

^{2}

*b*))+Tr(-

*b*ζ

^{3}/(

*a*+ζ

^{3}

*b*))+...+Tr(-

*b*ζ

^{p-1}/(

*a*+ζ

^{p-1}

*b*))=

*b*(

*a*

^{p}^{-2}-2

*a*

^{p}^{-3}

*b*+3

*a*

^{p}^{-4}

*b*

^{2}-...-(

*p*-1)

*b*

^{p}^{-2})/((

*a*)/(

^{p}+b^{p}*a+b*)), therefore if

*p*divides

*a+b*and (

*a*)/(

^{p}+b^{p}*a+b*)/

*p*is a

*p*th power, then {

*q*

^{p}^{-1}/[(

*a+b*)/(

*a*+ζ

^{p-1}

*b*)]}=

*ζ*where

^{uw}*w*=

*b*(

*a*

^{p}^{-2}-2

*a*

^{p}^{-3}

*b*+3

*a*

^{p}^{-4}

*b*

^{2}-...-(

*p*-1)

*b*

^{p}^{-2})/((

*a*)/(

^{p}+b^{p}*a+b*)).

*a*

^{p}^{-2}-2

*a*

^{p}^{-3}

*b*+3

*a*

^{p}^{-4}

*b*

^{2}-...-(

*p*-1)

*b*

^{p}^{-2}is congruent to (

*a+b*)

^{p-2}modulo

*p*, therefore 1/

*p*does not divide

*w*. By Proposition (25), if

*a*=

^{p}+b^{p}*c*,

^{p}*p*divides

*c*, then

*b*

^{p}^{-1}≡1(mod

*p*

^{2}) and hence {

*b*

^{p}^{-1}/[(

*a+b*)/(

*a*+ζ

^{p-1}

*b*)]}=1. The following proposition is based on data collected for

*p*=3;

(26) If 2 divides

*b*,

*p*divides

*a+b*, and (

*a*)/(

^{p}+b^{p}*a+b*)/

*p*is a

*p*th power, then

*p*

^{2}divides (

*p*-1)

*a*

^{p}^{-2}-(

*p*-2)

*a*

^{p}^{-3}

*b*+(

*p*-3)

*a*

^{p}^{-4}

*b*

^{2}-...-

*b*

^{p}^{-2}and

*p*

^{3}divides

*b*

^{p}^{-2}-2

*b*

^{p}^{-3}

*a*+3

*b*

^{p}^{-4}

*a*

^{2}-...-(

*p*-1)

*a*

^{p}^{-2}. If 2

*p*divides

*a+b*and (

*a*)/(

^{p}+b^{p}*a+b*)/

*p*is a

*p*th power, then

*p*

^{2}does not divide (

*p*-1)

*a*

^{p}^{-2}-(

*p*-2)

*a*

^{p}^{-3}

*b*+(

*p*-3)

*a*

^{p}^{-4}

*b*

^{2}-...-

*b*

^{p}^{-2}or

*b*

^{p}^{-2}-2

*b*

^{p}^{-3}

*a*+3

*b*

^{p}^{-4}

*a*

^{2}-...-(

*p*-1)

*a*

^{p}^{-2}.

#### Vandiver's Theorem

In 1919, Vandiver proved that if*a*=

^{p}+b^{p}*c*,

^{p}*p*divides

*c*, then

*p*

^{3}divides

*c*,

*a*

^{p}^{-1}≡1(mod

*p*

^{3}), and

*b*

^{p}^{-1}≡1(mod

*p*

^{3}). When

*a*is odd, Vandiver's theorem gives a necessary condition for a factor of

*a*to be a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*) (based on data collected for

*p*=3). Analogous results hold for

*b*,

*a-b*, and

*a+b*. The following proposition is based on data collected for

*p*=3;

(27) If [(

*a*)/(

^{p}+b^{p}*a+b*)]=

*T*and

^{p}*T*=

*U*where

^{k}*U*is a prime and

*p*does not divide

*k*, 2

*p*divides

*a*,

*b*,

*a-b*, or

*a+b*, 2 does not divide

*a*,

*q*divides

*a*, and

*q*

^{p}^{-1}≡1(mod

*p*

^{3}), then

*q*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*). If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power, 2

*p*divides

*a*,

*b*,

*a-b*, or

*a+b*, 2 does not divide

*a*, and

*q*divides

*a*, then

*q*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*) only if

*q*

^{p}^{-1}≡1(mod

*p*

^{3}). Analogous results hold for

*b*,

*a-b*, and

*a+b*. If [(

*a*)/(

^{p}+b^{p}*a+b*)]=

*T*and

^{p}*T*=

*U*where

^{k}*U*is a prime and

*p*does not divide

*k*, 2 divides

*a*,

*p*does not divide

*a*,

*q*divides

*a*, and every prime factor of

*q*is a

*p*th power modulo

*p*

^{2}, then

*q*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*). Analogous results hold for

*b*. If [(

*a*)/(

^{p}+b^{p}*a+b*)]=

*T*and

^{p}*T*=

*U*where

^{k}*U*is a prime and

*p*does not divide

*k*,

*p*divides

*a+b*, 2 does not divide

*a+b*,

*q*divides

*a+b*, and

*q*

^{p}^{-1}≡1(mod

*p*

^{3}), then

*q*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*). If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power,

*p*divides

*a+b*, 2 does not divide

*a+b*,

*q*divides

*a+b*, and

*p*does not divide

*q*, then

*q*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*) only if

*q*

^{p}^{-1}≡1(mod

*p*

^{3}).

####
Euler's Theorem and "Split" 2 and
*p*

Euler proved that every prime of the
form 6*k*+1 can be represented by

*x*

^{2}+3

*y*

^{2}. Let

*T*be a natural number and

*x*,

*y*, and

*z*be integers. If

*p*=3, every prime factor of

*T*is of the form 6

*k*+1, and

*T*has

*n*such distinct prime factors, then

*T*or

^{p}*pT*has exactly

^{p}*pn*representations of the form (

*a*)/(

^{p}+b^{p}*a+b*). All representations of

*pT*must be of the same type, that is, if (

^{p}*a*)/(

^{p}+b^{p}*a+b*) is one representation, then

*p*divides

*a+b*, and if ((

*a*')

^{p}+(

*b*')

^{p})/(

*a*'+

*b*') is another representation, then

*p*must divide

*a*'+

*b*'. Representations of

*T*can be of different types, that is,

^{p}*p*can divide

*a*,

*b*, or

*a-b*. Suppose

*p*=3, (

*a*)/(

^{p}+b^{p}*a+b*) is a representation of

*pT*, and 2 and

^{p}*p*are common factors of

*a+b*. When

*p*=3 and

*x+y*=z, (

*x*)/(

^{p}-y^{p}*x-y*)=(

*z*)/(

^{p}+y^{p}*z+y*)=(

*z*)/(

^{p}+x^{p}*z+x*), so 2 must divide

*b*' where

*b*'=

*a-b*and

*a*'=

*a*for the representation ((

*a*')

^{p}+(

*b*')

^{p})/(

*a*'+

*b*') of

*pT*(and

^{p}*p*must divide

*a*'+

*b*') and 2 must divide

*a*'' where

*a*''=

*a-b*and

*b*''=-

*b*for the representation ((

*a*'')

^{p}+(

*b*'')

^{p})/(

*a*''+

*b*'') of

*pT*(and

^{p}*p*must divide

*a*''+

*b*'').

#### A Generalization of Vandiver's Theorem

There is some evidence that if there is one representation [(*a*)/(

^{p}+b^{p}*a+b*)] of

*T*for

^{p}*p*>3, there must be other representations. If

*a*=

^{p}+b^{p}*c*,

^{p}*p*divides

*c*, and every prime factor of (

*c*)/(

^{p}-b^{p}*c-b*) and (

*c*)/(

^{p}-a^{p}*c-a*) is of the form

*p*

^{2}

*k*+1, then

*p*

^{3}divides

*c*by Barlow's formulas (since

*a*+

^{p}+b^{p}*a+b*-2

*c*=

*r*(

^{p}*R*-1)+

^{p}*s*(

^{p}*S*-1)). Vandiver's theorem suggests that no prime factor of (

^{p}*c*)/(

^{p}-b^{p}*c-b*) or (

*c*)/(

^{p}-a^{p}*c-a*) can just be of the form

*pk*+1. Vandiver's theorem can be reformulated so that it is applicable to the problem of determining if [(

*a*)/(

^{p}+b^{p}*a+b*)] can be a

*p*th power. The following proposition is based on data collected for

*p*=3;

(28) If [(

*a*)/(

^{p}+b^{p}*a+b*)]=

*T*where every prime factor of

^{p}*T*is of the form

*p*

^{2}

*k*+1,

*p*

^{3}divides

*a*,

*b*, or

*a-b*or

*p*

^{4}divides

*a+b*, and 2 does not divide

*a*, then

*a*

^{p}^{-1}≡1(mod

*p*

^{3}). If [(

*a*)/(

^{p}+b^{p}*a+b*)]=

*T*where

^{p}*T*has only one distinct prime factor, this prime factor is of the form

*p*

^{2}

*k*+1,

*p*

^{3}divides

*a*,

*b*, or

*a-b*or

*p*

^{4}divides

*a+b*or

*p*

^{3}divides

*a*',

*b*', or

*a*'-

*b*' or

*p*

^{4}divides

*a*'+

*b*' for some representation [((

*a*')

^{p}+(

*b*')

^{p})/(

*a*'+

*b*')] of

*T*, and 2 does not divide

^{p}*a*, then

*a*

^{p}^{-1}≡1(mod

*p*

^{3}). Analogous results hold for

*b*and

*a-b*. If [(

*a*)/(

^{p}+b^{p}*a+b*)]=

*T*where every prime factor of

^{p}*T*is of the form

*p*

^{2}

*k*+1,

*p*

^{3}divides

*a*,

*b*, or

*a-b*or

*p*

^{4}divides

*a+b*, and 2 does not divide

*a+b*, then [(

*a+b*)/

*p*]

^{p-1}≡1(mod

*p*

^{3}) if

*p*divides

*a+b*, or (

*a+b*)

^{p-1}≡1(mod

*p*

^{3}) if

*p*does not divide

*a+b*. If [(

*a*)/(

^{p}+b^{p}*a+b*)]=

*T*where

^{p}*T*has only one distinct prime factor, this prime factor is of the form

*p*

^{2}

*k*+1,

*p*

^{3}divides

*a*,

*b*, or

*a-b*or

*p*

^{4}divides

*a+b*or

*p*

^{3}divides

*a*',

*b*', or

*a*'-

*b*' or

*p*

^{4}divides

*a*'+

*b*' for some representation [((

*a*')

^{p}+(

*b*')

^{p})/(

*a*'+

*b*')] of

*T*, and 2 does not divide

^{p}*a+b*, then [(

*a+b*)/

*p*]

^{p-1}≡1(mod

*p*

^{3}) if

*p*divides

*a+b*, or (

*a+b*)

^{p-1}≡1(mod

*p*

^{3}) if

*p*does not divide

*a+b*. If [(

*a*)/(

^{p}+b^{p}*a+b*)]=

*T*where every prime factor of

^{p}*T*is of the form

*p*

^{2}

*k*+1, 2 divides

*a*, and

*p*does not divide

*a*, then (

*a*/2)

^{p-1}≡1(mod

*p*

^{3}). Analogous results hold for

*b*. If [(

*a*)/(

^{p}+b^{p}*a+b*)]=

*T*where every prime factor of

^{p}*T*is of the form

*p*

^{2}

*k*+1, 2 does not divide

*a*, and

*p*divides

*a+b*, then

*a*

^{p}^{-1}≡1(mod

*p*

^{3}). Analogous results hold for

*b*and

*a-b*.

#### Congruence Properties of
Prime Factors of [(*a*^{p}+b^{p})/(*a+b*)]
when [(*a*^{p}+b^{p})/(*a+b*)]
is a *p*th Power

That the reformulation of Vandiver's
theorem depends on different
representations [(^{p}+b

^{p}

^{p}+b

^{p}

*a*)/(

^{p}+b^{p}*a+b*)] of

*T*is some indication that different representations must exist for

^{p}*p*>3 (if there are any representations). Whether

*p*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*) is of importance to Vandiver's theorem. The following propositions are based on data collected for

*p*=3;

(29) If [(

*a*)/(

^{p}+b^{p}*a+b*)]=

*T*and

^{p}*T*=

*U*where

^{k}*U*is a prime and

*p*does not divide

*k*, then

*p*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*) if and only if

*p*

^{3}divides

*a*,

*b*, or

*a-b*or

*p*

^{4}divides

*a+b*or

*p*

^{3}divides

*a*',

*b*', or

*a*'-

*b*' or

*p*

^{4}divides

*a*'+

*b*' for some representation [((

*a*')

^{p}+(

*b*')

^{p})/(

*a*'+

*b*')] of

*T*. If [(

^{p}*a*)/(

^{p}+b^{p}*a+b*)]=

*T*and

^{p}*T*=

*U*where

^{k}*U*is a prime and

*p*divides

*k*, then

*p*is not a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*). If [(

*a*)/(

^{p}+b^{p}*a+b*)]=

*T*where

^{p}*T*has two distinct prime factors, 2

*p*divides

*a*,

*b*,

*a-b*, or

*a+b*, and

*p*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*), then

*p*

^{3}divides

*a*,

*b*, or

*a-b*or

*p*

^{4}divides

*a+b*.

(30) If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power, then

*p*

^{p}^{-1}

*a*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*) if 2

*p*divides

*a*,

*p*

^{p}^{-1}

*b*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*) if 2

*p*divides

*b*,

*p*

^{p}^{-1}(

*a-b*) and

*a+b*are

*p*th powers w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*) if 2

*p*divides

*a-b*, or

*a-b*and

*p*(

*a+b*) are

*p*th powers w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*) if 2

*p*divides

*a+b*.

(31) If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power, 2 divides

*a*, and

*p*does not divide

*a*,

*then*

*a*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*). Analogous results hold for

*b*.

(32) If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power,

*f*is a prime factor of [(

*a*)/(

^{p}+b^{p}*a+b*)] of the form

*p*

^{2}

*k*+1, and

*p*is not a

*p*th power modulo

*f*, then (1)

*p*

^{p}^{-1}

*a*,

*p*

^{p}^{-1}

*b*,

*p*

^{p}^{-1}(

*a-b*), and

*a+b*are

*p*th powers modulo

*f*if

*p*divides

*a*,

*b*, or

*a-b*, or (2)

*a*,

*b*,

*a-b*, and

*p*(

*a+b*) are

*p*th powers modulo

*f*if

*p*divides

*a+b*. If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power,

*f*is a prime factor of [(

*a*)/(

^{p}+b^{p}*a+b*)] of the form

*p*

^{2}

*k*+1, and

*p*is a

*p*th power modulo

*f*, then

*a*,

*b*,

*a-b*, and

*a+b*are

*p*th powers modulo

*f*.

(33) If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power,

*f*is a prime factor of [(

*a*)/(

^{p}+b^{p}*a+b*)] not of the form

*p*

^{2}

*k*+1, and

*p*is not a

*p*th power modulo

*f*, then (1)

*p*

^{p}^{-1}

*a*,

*b*,

*p*(

*a-b*), and

*p*

^{p}^{-1}(

*a+b*), or

*p*

^{p}^{-1}

*a*,

*pb*,

*a-b*, and

*p*(

*a+b*) are

*p*th powers modulo

*f*if 2

*p*divides

*a*, or (2)

*a*,

*p*

^{p}^{-1}

*b*,

*p*(

*a-b*), and

*p*

^{p}^{-1}(

*a+b*), or

*pa*,

*p*

^{p}^{-1}

*b*,

*a-b*, and

*p*(

*a+b*) are

*p*th powers modulo

*f*if 2

*p*divides

*b*, or (3)

*a*,

*pb*,

*p*

^{p}^{-1}(

*a-b*), and

*a+b*, or

*pa*,

*b*,

*p*

^{p}^{-1}(

*a-b*), and

*a+b*are

*p*th powers modulo

*f*if 2

*p*divides

*a-b*, or (4)

*pa*,

*p*

^{p}^{-1}

*b*,

*a-b*, and

*p*(

*a+b*), or

*p*

^{p}^{-1}

*a*,

*pb*,

*a-b*, and

*p*(

*a+b*) are

*p*th powers modulo

*f*if 2

*p*divides

*a+b*.

(34) If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power, 2 divides

*a*,

*p*does not divide

*a*,

*f*is a prime factor of [(

*a*)/(

^{p}+b^{p}*a+b*)] not of the form

*p*

^{2}

*k*+1, and

*p*is not a

*p*th power modulo

*f*, then

*a*,

*pb*,

*p*

^{p}^{-1}(

*a-b*), and

*a+b*, or

*a*,

*p*

^{p}^{-1}

*b*,

*p*(

*a-b*), and

*p*

^{p}^{-1}(

*a+b*) are

*p*th powers modulo

*f*. Analogous results hold for

*b*.

Since Propositions (30), (32), (33), and (34) are based solely on data collected for

*p*=3, their form is ambiguous in that the

*p*exponents might be 2 instead of

*p*-1. Congruence properties of the prime factors of [(

*a*)/(

^{p}-b^{p}*a-b*)] when [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power appear to determine the form of Propositions (30), (32), (33), and (34). (Propositions (30), (32), (33), and (34) can be transformed into Propositions (12), (8), (14), and (9) respectively by multiplying the

*a*,

*b*,

*a-b*, and

*a+b*terms by

*p*and switching the

*a+b*and

*a-b*terms [and of course the moduli bases]. This is just an attempt to find a simple relationship between the congruence properties of the prime factors of [(

*a*)/(

^{p}+b^{p}*a+b*)] and [(

*a*)/(

^{p}-b^{p}*a-b*)] when [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power and has no apparent logical basis.)

(35) If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power,

*f*is a prime factor of [(

*a*)/(

^{p}+b^{p}*a+b*)] not of the form

*p*

^{2}

*k*+1, and

*p*is a

*p*th power modulo

*f*, then (1)

*a*(and not

*b*,

*a-b*, or

*a+b*) is a

*p*th power modulo

*f*if 2 divides

*a*, or (2)

*b*(and not

*a*,

*a-b*, or

*a+b*) is a

*p*th power modulo

*f*if 2 divides

*b*, or (3)

*a-b*and

*a+b*(and not

*a*or

*b*) are

*p*th powers modulo

*f*if 2 divides

*a-b*or

*a+b*.

As shown previously,

*a*=

^{p}+b^{p}*c*,

^{p}*p*divides

*c*, implies

*c*,

*pb*, and

*c-b*are

*p*th powers w.r.t. (

*c*)/(

^{p}-b^{p}*c-b*),

*c*,

*pa*, and

*c-a*are

*p*th powers w.r.t. (

*c*)/(

^{p}-a^{p}*c-a*), and 2 and

*p*are common factors of

*c*(if the reformulated version of Furtwängler's theorem is accepted). Then by Propositions (33) and (35), every prime factor of (

*c*)/(

^{p}-b^{p}*c-b*) and (

*c*)/(

^{p}-a^{p}*c-a*) must be of the form

*p*

^{2}

*k*+1. (Substituting

*c*for

*a*and -

*b*for

*b*in Proposition (33) gives

*p*

^{p}^{-1}

*c*, -

*b*,

*p*(

*c+b*), and

*p*

^{p}^{-1}(

*c-b*), or

*p*

^{p}^{-1}

*c*, -

*pb*,

*c+b*, and

*p*(

*c-b*) are

*p*th powers modulo

*f*[a prime factor of (

*c*)/(

^{p}-b^{p}*c*-

*b*)] if 2

*p*divides

*c*and

*p*is not a

*p*th power modulo

*f*[a contradiction]. Substituting

*c*for

*a*and -

*b*for

*b*in Proposition (35) gives

*c*[and not

*c-b*] is a

*p*th power modulo

*f*if 2

*p*divides

*c*and

*p*is a

*p*th power modulo

*f*[a contradiction]. Analogous results follow by substituting

*c*for

*a*and -

*a*for

*b*in Propositions (33) and (35).) Furthermore, by Proposition (32),

*p*must be a

*p*th power w.r.t. (

*c*)/(

^{p}-b^{p}*c-b*) and (

*c*)/(

^{p}-a^{p}*c-a*). As shown previously,

*a*=

^{p}+b^{p}*c*,

^{p}*p*divides

*c*, implies

*a*,

*b*, and

*p*(

*a+b*) are

*p*th powers w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*), and every prime factor of [(

*a*)/(

^{p}+b^{p}*a+b*)] is of the form

*p*

^{2}

*k*+1. This gives the following proposition;

(36) If

*a*=

^{p}+b^{p}*c*where

^{p}*p*divides

*c*, then every prime factor of (

*c*)/(

^{p}-b^{p}*c-b*) is of the form

*p*

^{2}

*k*+1 and

*c*,

*b*,

*c-b*,

*c+b*, and

*p*are

*p*th powers w.r.t. (

*c*)/(

^{p}-b^{p}*c-b*). Analogous results hold for (

*c*)/(

^{p}-a^{p}*c-a*). If

*a*=

^{p}+b^{p}*c*where

^{p}*p*divides

*c*, then every prime factor of [(

*a*)/(

^{p}+b^{p}*a+b*)] is of the form

*p*

^{2}

*k*+1 and

*a*,

*b*,

*a-b*, and

*p*(

*a+b*) are

*p*th powers w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*).

More evidence for the above proposition is given by the following three propositions (based on data collected for

*p*=3, 5, and 7);

(37) If every prime factor of [(

*a*)/(

^{p}+b^{p}*a+b*)] is of the form

*p*

^{2}

*k*+1 and

*p*

^{2}divides

*a*,

*b*,

*a-b*, or

*a+b*, then

*a*

^{p}^{-1}≡1(mod

*p*

^{2}) if

*p*does not divide

*a*,

*b*

^{p}^{-1}≡1(mod

*p*

^{2}) if

*p*does not divide

*b*, and (

*a-b*)

^{p-1}≡1(mod

*p*

^{2}) and (

*a+b*)

^{p-1}≡1(mod

*p*

^{2}) if

*p*does not divide

*a-b*or

*a+b*.

(38) If

*a*,

*pb*, and

*a+b*are

*p*th powers w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*) and

*p*

^{2}divides

*a*, then

*b*,

*a+b*, and

*a-b*are

*p*th powers modulo

*p*

^{2}. If

*p*=3 or 5,

*a*,

*pb*, and

*a+b*are

*p*th powers w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*), and

*p*

^{2}divides

*a*,

*b*,

*a-b*, or

*a+b*, then

*a-b*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*) if and only if

*p*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*).

*(39) If*

*a*,

*b*,

*and*

*p*(

*a+b*) are

*p*th powers w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*) and

*p*

^{2}divides

*a*or

*b*, then

*a+b*and

*a-b*are

*p*th powers modulo

*p*

^{2}.

*If*

*p*=3,

*a*,

*b*, and

*p*(

*a+b*) are

*p*th powers w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*), and

*p*

^{2}divides

*a*,

*b*,

*a-b*, or

*a+b*, then

*a-b*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*).

Substituting

*c*for

*a*and -

*b*for

*b*in Proposition (14) gives

*p*

^{2}(

*c-b*) is a

*p*th power modulo

*f*(a prime factor of (

*c*)/(

^{p}+b^{p}*c+b*) not of the form

*p*

^{2}

*k*+1) if 2

*p*divides

*c*and

*p*/2 is a

*p*th power modulo

*f*, or

*p*(

*c-b*) is a

*p*th power modulo

*f*if 2

*p*divides

*c*and

*2*

*p*is a

*p*th power modulo

*f*. Then if

*c-b*is a

*p*th power,

*p*must be a

*p*th power modulo

*f*and hence by Proposition (7), 2 cannot be a

*p*th power modulo

*f*(otherwise, 2

*p*would be a

*p*th power modulo

*f*). As shown previously,

*a*=

^{p}+b^{p}*c*implies 2 is a

^{p}*p*th power w.r.t. (

*c*)/(

^{p}+b^{p}*c+b*). This gives the following proposition;

(40) If

*a*=

^{p}+b^{p}*c*where

^{p}*p*divides

*c*, then every prime factor of (

*c*)/(

^{p}+b^{p}*c+b*) is of the form

*p*

^{2}

*k*+1 and

*c*,

*b*,

*p*(

*c+b*) and

*c-b*are

*p*th powers w.r.t. (

*c*)/(

^{p}+b^{p}*c+b*). Analogous results hold for (

*c*)/(

^{p}+a^{p}*c+a*).

The following proposition is based on data collected for

*p*=3;

(41) If [(

*a*)/(

^{p}+b^{p}*a+b*)] is a

*p*th power and

*f*is a prime factor of [(

*a*)/(

^{p}+b^{p}*a+b*)], then at least one of 2

*p*, 2,

*p*, or

*p*/2 is a

*p*th power modulo

*f*.

The following propositions are based on data collected for

*p*=3, 5, and 7;

(42) If

*p*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*), then

*p*

^{2}divides

*a*if

*p*divides

*a*,

*p*

^{2}divides

*b*if

*p*divides

*b*, or

*p*

^{2}divides

*a-b*if

*p*divides

*a-b*. If

*p*>3 and

*p*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*), then

*p*

^{2}divides

*a+b*if

*p*divides

*a+b*. If

*p*=3 and

*p*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*),

*then*

*p*

^{3}divides

*a+b*if

*p*

^{2}divides

*a+b*.

When

*p*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*) and [(

*a*)/(

^{p}+b^{p}*a+b*)] is not prime, the "small" prime factors of [(

*a*)/(

^{p}+b^{p}*a+b*)] are not of the form

*p*

^{2}

*k*+1. For example, of the 2517 prime factors (not necessarily distinct) of [(

*a*)/(

^{p}+b^{p}*a+b*)] for the 1175 (

*a*,

*b*) such that

*p*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*), [(

*a*)/(

^{p}+b^{p}*a+b*)] is not prime, and 1000≥

*a*>

*b*≥1 for

*p*=7, only 214 prime factors are of the form

*p*

^{2}

*k*+1 and the smallest of these prime factors is 15877. When

*p*=3 and

*p*is a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*), the three smallest prime factors of [(

*a*)/(

^{p}+b^{p}*a+b*)] of the form

*p*

^{2}

*k*+1 are 73, 271, and 307.

(43) If

*p*=3,

*a*,

*b*,

*a-b*, and

*a+b*are

*p*th powers w.r.t. (

*a*)/(

^{p}+b^{p}*a*+

*b*),

*p*

^{2}divides

*a*,

*b*,

*a-b*, or

*a+b*, and [(

*a*)/(

^{p}+b^{p}*a+b*)] is not prime, then every prime factor of [(

*a*)/(

^{p}+b^{p}*a+b*)] equals [((

*a*')

^{p}+(

*b*')

^{p})/(

*a*'+

*b*')] where

*p*

^{2}divides

*a*',

*b*',

*a*'-

*b*', or

*a*'+

*b*'. (The smallest possible prime factor of [(

*a*)/(

^{p}+b^{p}*a+b*)] satisfying these conditions is 73; the requirement that

*p*

^{2}divides

*a*',

*b*',

*a*'-

*b*', or

*a*'+

*b*' eliminates about 2/3 of the primes of the form

*p*

^{2}

*k+*1 from consideration.)

*For the 13208764 (*

*a*,

*b*) such that 50000≥

*a*>

*b*≥1,

*a*,

*b*,

*a-b*, and

*a+b*are

*p*th powers w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*), and

*p*

^{2}divides

*a*,

*b*,

*a-b*, or

*a+b*for

*p*=3, the numbers of instances where [(

*a*)/(

^{p}+b^{p}*a+b*)] has 1, 2, 3, and 4 prime factors (not necessarily distinct) are 12585008, 615167, 8518, and 71 respectively. [(

*a*)/(

^{p}+b^{p}*a+b*)] is a square in 624 instances, a cube in 27 instances, and a fourth power in 3 instances. For larger upper bounds of the

*a*,

*b*values, the proportions of the numbers of instances where [(

*a*)/(

^{p}+b^{p}*a+b*)] has 2, 3, and 4 prime factors increase, so there should eventually be a value of [(

*a*)/(

^{p}+b^{p}*a+b*)] having 5 or more prime factors.

(44) If

*p*=3,

*a*,

*b*,

*a-b*, and

*p*(

*a+b*) are

*p*th powers w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*),

*p*is not a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*),

*p*

^{2}divides

*a*,

*b*,

*a-b*, or

*a+b*, and [(

*a*)/(

^{p}+b^{p}*a*+

*b*)] is not prime, then every prime factor of [(

*a*)/(

^{p}+b^{p}*a+b*)] equals [((

*a*')

^{p}+(

*b*')

^{p})/(

*a*'+

*b*')] where p

^{2}does not divide

*a*',

*b*',

*a*'-

*b*', or

*a*'+

*b*'.

For the 1316973 (

*a*,

*b*) such that 25000≥

*a*>

*b*≥1,

*a*,

*b*,

*a-b*, and

*p*(

*a+b*) are

*p*th powers w.r.t.(

*a*)/(

^{p}+b^{p}*a+b*),

*p*is not a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*), and

*p*

^{2}divides

*a*,

*b*,

*a-b*, or

*a+b*for

*p*=3, the numbers of instances where [(

*a*)/(

^{p}+b^{p}*a+b*)] has 1, 2, 3, 4, 5, and 6 prime factors are 712815, 573912, 29149, 1002, 88, and 7 respectively. [(

*a*)/(

^{p}+b^{p}*a+b*)] is a square in 112 instances, a cube in 9 instances, and a fourth power in 2 instances.

If

*p*=5, 5000≥

*a>b*≥1,

*a*,

*b*,

*a-b*, and

*a+b*are

*p*th powers w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*), and

*p*

^{2}divides

*a*,

*b*,

*a-b*, or

*a+b*, then [(

*a*)/(

^{p}+b^{p}*a+b*)] has at most two prime factors. If

*p*=7, 500≥

*a*>

*b*≥1,

*a*,

*b*,

*a-b*, and

*a+b*are

*p*th powers w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*), and

*p*

^{2}divides

*a*,

*b*,

*a-b*, or

*a+b*, then [(

*a*)/(

^{p}+b^{p}*a+b*)] is prime. If

*p*=5, 5000≥

*a>b*≥1,

*a*,

*b*,

*a-b*, and

*p*(

*a+b*) are

*p*th powers w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*),

*p*is not a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*),

*and*

*p*

^{2}divides

*a*,

*b*,

*a-b*, or

*a+b*, then [(

*a*)/(

^{p}+b^{p}*a+b*)] has at most three prime factors. If

*p*=7 and 500≥

*a*>

*b*≥1, there do not exist (

*a*,

*b*) such that

*a*,

*b*,

*a-b*, and

*p*(

*a+b*) are

*p*th powers w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*),

*p*is not a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*),

*and*

*p*

^{2}divides

*a*,

*b*,

*a-b*, or

*a+b*. For

*p*=5, the prime factors of [(

*a*)/(

^{p}+b^{p}*a+b*)] when

*a*,

*b*,

*a-b*,

*a+b*are

*p*th powers w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*) and

*p*

^{2}divides

*a*,

*b*,

*a-b*, or

*a+b*are different from the prime factors of [(

*a*)/(

^{p}+b^{p}*a+b*)] when

*a*,

*b*,

*a-b*, and

*p*(

*a+b*) are

*p*th powers w.r.t. (

*a*),

^{p}+b^{p}*p*is not a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*), and

*p*

^{2}divides

*a*,

*b*,

*a-b*, or

*a+b*, the same as for

*p*=3. For

*p*=3, this was due to the representations of the prime factors. This is some indication that representations (of the form [(

*a*)/(

^{p}+b^{p}*a+b*)]) of the prime factors are still relevant for

*p*>3. A table giving the factorization of [(

*a*)/(

^{p}+b*a+b*)] when

*a*,

*b*,

*a-b*, and

*a+b*are

*p*th powers w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*) and

*p*

^{2}divides

*a*,

*b*,

*a-b*, or

*a+b*for

*p*=5

*is out43p5. For*

*p*=5 and 5000≥

*a*>

*b*≥1,

*there are 25287 prime values of [(*

*a*)/(

^{p}+b^{p}*a+b*)] where the (

*a*,

*b*) values satisfy these conditions. A table giving the factorization of [(

*a*)/(

^{p}+b*a+b*)] when

*a*,

*b*,

*a-b*, and

*p*(

*a+b*) are

*p*th powers w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*),

*p*is not a

*p*th power w.r.t. (

*a*)/(

^{p}+b^{p}*a+b*),

*and*

*p*

^{2}divides

*a*,

*b*,

*a-b*, or

*a+b*for

*p*=5

*is out44p5. There are no prime values of [(*

*a*)/(

^{p}+b^{p}*a+b*)] where the (

*a*,

*b*) values satisfy these conditions.

#### References

1810 Barlow, P.*Demonstration of a curious numerical proposition*.

*J. Nat. Phil. Chem. and Arts, 27, 1810, 193-205.*

1909 Wieferich, A.

*Zum Letzten Fermat'schen Theorem*.

*J. reine u. angew. Math. 136, 1909, 293-302.*

1911 Mirimanoff, D.

*Sur le dernier théorème de Fermat.*J. reine u. angew. Math., 139, 1911, 309-324.

1912 Furtwängler, P.

*Letzter Fermatschen Satz und Eisensteins'ches Reziprozitätsgesetz*. Sitzungsber, Akad. d. Wiss. Wein., Abt. IIa, 121, 1912, 589-592.

1919 Vandiver, H. S.

*A property of cyclotomic integers and its relation to Fermat's last theorem*. Annals of Math., 21, 1919, 73-80.

1970 Hasse, H.

*Bericht über neuere Undersuchungen und Probleme aus der Theorie der algebraischen Zahlkörper*. Teil I: Klassenkörperietheorie. Teil Ia: Beweise zu Tiel I; Teil II: Reziprozitätsgesetz. 3. Aufl.. Wurzburg Wien: Physika-Verlag. 204 pp. Jbuch 52-150; 53, 143; 46-165