Sophie Germain (1776-1831)

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: xp + yp = zp 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: xp + yp = zp 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 q with the properties that • xp = p mod q is impossible for any value of x • the equation r’= r+1 mod q cannot be satisfied for any pth 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 xp + yp = zp implies that x = 0 mod q, or y = 0 mod q, or z = 0 mod q. • Proof: If the equation would hold, and say x is not 0 mod q, we can multiply by ap where a is the inverse of x mod q. Then 1+(ay)p(az)p gives consecutive pth powers. • Proof of Theorem: Step 1: Factorize xp+yp=(x+y)f(x,y) with f(x,y)= xp-1-xp-2y+xp-3y2-….+yp-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)=pxp-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 pth powers. • Set x+y=lp and f(x,y)=rp, so z=lr. • In the same way one gets equations z-y=hp and z-x=vp Step 3: By the Basic Lemma either x,y or z is a multiple of q. Say z  0 mod q. Then lp+hp+vp =2z 0 mod q Also one of the l,h,v has to be divisible byq. (Same argument as in the Lemma). Step 4: Since we are looking for primitive solutions only l can be divisible by q. If h would be then q would be a common factor of z and y. and if v where then q would be a common factor of x and z. So x+y=lp 0 mod q sox -y mod q and rp pxp-1 p(-vp)p-1 pvp(p-1) mod q, so p (r/vp-1)p mod q, which contradicts the assumption 1.