Transcript Sophie Germain’s Theorem

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Sophie Germain
(1776-1831)
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• 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.
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.
Sophie Germain’s Theorem
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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
1. xp = p mod q is impossible for any value of x
2. 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.
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.
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 by q. (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 so x  -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.