Upload
kathleen-tanner
View
217
Download
0
Embed Size (px)
DESCRIPTION
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 - PowerPoint PPT Presentation
Citation preview
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.
QuickTime™ and aTIFF (Uncompressed) decompressor
are needed to see this picture.
QuickTime™ and aTIFF (Uncompressed) decompressor
are needed to see this picture.
Congruences
• Let p be a prime.• For the residues modulo p there is addition, subtraction, multiplication
and division.• If a=kp+r we write ar mod p, or ar (p).
• Notice that if a=kp+r and b=lp+s, then a+b r+s mod p.
• Also -rp-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 rs0 (p) then rs=kp and either r0
(p) or s0 (p). This means that we can invert multiplication to division.
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 with the properties that
1. xp = p mod is impossible for any value of x2. the equation r’= r+1 mod cannot be satisfied for any pth powers
then Case I of Fermat's Last Theorem is true for p.
Sophie Germain’s Theorem
• Basic Lemma: If the condition 2 holds then xp + yp = zp 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 ap where a is the inverse of x mod . 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.
Sophie Germain’s TheoremStep 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 . Say z 0 mod Then
lp+hp+vp =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=lp 0 mod sox -y mod and
rp pxp-1 p(-vp)p-1 pvp(p-1) mod so
p (r/vp-1)p mod , which contradicts the assumption 1.