1. Sophie Germain (1776-1831) Marie-Sophie Germain, studied independently using lecture notes for many courses from École Polytechnique. She was supported by her parents She used the pseudonym M. LeBlanc Corresponded with Lagrange who found out she was a woman and supported her. He also put several discoveries of hers as a supplement in his book Essai sur le Théorie des Nombres. Corresponded with Gauss who had high esteem for her work in number theory. Most famous result on the elasticity of surfaces to explain Chladni figures, for which she was awarded a prize. Also worked on FLT; a letter to Gauss in 1819 outlines her strategy. Sophie Germain’s Theorem was published in a footnote of Legedre’s 1827 Memoir in which he proves FLT for n=5.

2. Congruences Let p be a prime. For the residues modulo p there is addition, subtraction, multiplication and division. If a=kp+r we write a  r mod p, or a  r (p). Notice that if a=kp+r and b=lp+s, then a+b  r+s mod p. Also -r  p-r mod p, since if a=kp-r=(k-1)p+(p-r) and b=lp+r, then a+b  (k+l)p  0 mod p. Likewise we can multiply residues. Notice that since p is a prime, if rs  0 (p) then rs=kp and either r  0 (p) or s  0 (p). This means that we can invert multiplication to division.

3. Sophie Germain’s Theorem FLT I: x p + y p = z p has no integer solutions for which x, y, and z are relatively prime to p, i.e. in which none of x, y, and z are divisible by p; FLT II: x p + y p = z p has no integer solutions for which one and only one of the three numbers is divisible by p. Sophie Germain's Theorem: Let p be an odd prime. If there is an auxiliary prime  with the properties that 1.x p = p mod  is impossible for any value of x 2.the equation r’= r+1 mod  cannot be satisfied for any p th powers then Case I of Fermat's Last Theorem is true for p.

4. Sophie Germain’s Theorem Basic Lemma: If the condition 2 holds then x p + y p = z p implies that x = 0 mod , or y = 0 mod , or z = 0 mod  Proof: If the equation would hold, and say x is not 0 mod , we can multiply by a p where a is the inverse of x mod . Then 1+(ay) p  (az) p gives consecutive p th powers. Proof of Theorem: Step 1: Factorize x p +y p =(x+y)f(x,y) with f(x,y)= x p-1 -x p-2 y+x p-3 y 2 -….+y p-1 then (x+y) and f(x,y) are relatively prime. Assume this is not the case and q is a common prime divisor, then y=-x+kq and by substituting f(x,y)=px p-1 +rq. If p is divisible by q then p=q and x,y,z all are divisible by p which is a contradiction to the assumption. So x must be divisible by q and hence y. But x,y are coprime, so we get a contradiction.

5. Sophie Germain’s Theorem Step 2: By unique factorization x+y and f(x,y) both have to be p th powers. –Set x+y=l p and f(x,y)=r p, so z=lr. –In the same way one gets equations z-y=h p and z-x=v p Step 3: By the Basic Lemma either x,y or z is a multiple of . Say z  0 mod  Then l p +h p +v p =2z  0 mod  Also one of the l,h,v has to be divisible by . (Same argument as in the Lemma). Step 4: Since we are looking for primitive solutions only l can be divisible by . If h would be then  would be a common factor of z and y. and if v where then  would be a common factor of x and z. So x+y=l p  0 mod  so  x  -y mod  and r p  px p-1  p(-v p ) p-1  pv p(p-1) mod  so p  (r/v p-1 ) p mod , which contradicts the assumption 1.