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 Germains Theorem was published in a
  footnote of Legedres 1827 Memoir in which he
  proves FLT for n5.

2
Congruences

• Let p be a prime.
• For the residues modulo p there is addition,
  subtraction, multiplication and division.
• If akpr we write a?r mod p, or a?r (p).
• Notice that if akpr and blps, then ab ? rs
  mod p.
• Also -r?p-r mod p, since if akp-r(k-1)p(p-r)
  and blpr, then ab? (kl)p ? 0 mod p.
• Likewise we can multiply residues.
• Notice that since p is a prime, if rs?0 (p) then
  rskp and either r?0 (p) or s?0 (p). This means
  that we can invert multiplication to division.

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Sophie Germains 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 r1 mod q cannot be satisfied
  for any pth powers
• then Case I of Fermat's Last Theorem is true for
  p.

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Sophie Germains 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 xpyp(xy)f(x,y) with
• f(x,y) xp-1-xp-2yxp-3y2-.yp-1
• then (xy) and f(x,y) are relatively prime.
  Assume this is not the case and q is a common
  prime divisor, then y-xkq and by substituting
  f(x,y)pxp-1rq. If p is divisible by q then pq
  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 Germains Theorem

• Step 2 By unique factorization xy and f(x,y)
  both have to be pth powers.
• Set xylp and f(x,y)rp, so zlr.
• In the same way one gets equations
• z-yhp and z-xvp
• Step 3 By the Basic Lemma either x,y or z is a
  multiple of q. Say z ? 0 mod q. Then
• lphpvp 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
• xylp ? 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.