20.01.2015 Views

Sophie Germain: mathématicienne extraordinaire - Scripps College

Sophie Germain: mathématicienne extraordinaire - Scripps College

Sophie Germain: mathématicienne extraordinaire - Scripps College

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

<strong>Sophie</strong> <strong>Germain</strong>’s Theorems 52<br />

In other words, if there did exist any positive integer solution (a, b, c) of<br />

x n + y n = z n , then none of a, b, or c are multiples of n.<br />

5.2 <strong>Sophie</strong> <strong>Germain</strong>’s Theorems<br />

These two cases developed following <strong>Sophie</strong> <strong>Germain</strong>’s Theorem:<br />

<strong>Sophie</strong> <strong>Germain</strong>’s Theorem 1. Let p be an odd prime. Suppose there exists an<br />

auxiliary prime q such that the following conditions hold:<br />

1. n p ≢ p (mod q) for all integers n.<br />

2. Whenever x p + y p + z p ≡ 0 (mod q) then q|xyz where x, y, and z are<br />

relatively prime.<br />

Then for any non-zero integers a, b, and c such that a p + b p = c p , it must be that<br />

p|abc.<br />

That is, suppose a prime p has an auxiliary prime q such that conditions<br />

1 and 2 are met. If there exists non-zero integers a, b, and c for which<br />

a p + b p = c p , then p divides a, b, or c.<br />

Example.<br />

Let us consider the primes p = 3 and q = 7. We check directly<br />

conditions 1 and 2.<br />

1. Is n 3 ≡ 3 (mod 7) for any integer n We can easily check this with a<br />

table.

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!