Sophie Germain: mathématicienne extraordinaire - Scripps College

Sophie Germain’s Theorems 54
x 3 , y 3 , and z 3 are congruent to 0 (mod 7) or when only x 3 ≡ 0 (mod 7).
This implies that either x, y, and z were not relatively prime and
were in fact all divisible by seven or that x is divisible by 7. Thus
x 3 + y 3 + z 3 ≡ 0 (mod 7) implies that 7|xyz and condition 2 is met.
Since 3 is prime with an auxiliary prime 7 which fulfills conditions 1 and
2, we can conclude that if the equation x 3 + y 3 = z 3 has non-trivial integer
solutions x 0 , y 0 , and z 0 , then one of x 0 , y 0 , or z 0 is divisible by 3.
Proof of Sophie Germain’s Theorem.
Let p be a prime with auxiliary
prime q such that condition 2 is satisfied. Suppose x, y, and z are nonzero
integer solutions of the equation X p + Y p + Z p = 0. Suppose further
that p does not divide any of x, y, or z. We will show that if none of x, y, or
z is divisible by p then condition 1 is not satisfied.
For any integers x, y, and z, if x p +y p +z p = 0, then x p + y p + z p ≡ 0 (mod q).
By condition 2, we know that q|xyz.
Assume without loss of generality that q divides x. Then we can write
(−x) p = y p + z p . Factoring out y + z, we get
(−x) p = (y + z)(y p−1 − y p−2 z + y p−3 z 2 + · · · − yz p−2 + z p−1 ).
Recall the following lemma from chapter 4:
Lemma 1.
Relatively prime divisors of an nth power are themselves nth powers.
We want to show that y+z and y p−1 −y p−2 z+y p−3 z 2 +· · ·−yz p−2 +z p−1
are pth powers, so need to show that they are relatively prime.