Problem 35

ThePerfectHacker · #1 $Old$ 08-26-2007, 04:50 PM

The mathematician, Fermat, said that any prime $p$ of the form $4k+1$ can be expressed as a sum of two squares. However, it turns out amazingly, that this representation is also unique. Accept by faith the existence of Fermat's theorem and prove uniqueness.

SkyWatcher · #2 $Old$ 09-03-2007, 07:10 AM

I think I have got a solution, but's its also prooving the existence of the two numbers : no acceptation by faith!
would it fit?
(i hope not, because i feel lazy and though it's not very long it's a little tricky because it's involving the distinction betwenn numbers and their 'correspondant' modulo 4k+1=p')

ThePerfectHacker · #3 $Old$ 09-04-2007, 06:48 PM

Say that $p = a^2+b^2 = c^2 + d^2$ .

Note that $a^2d^2 - b^2c^2 = (p-b^2)d^2 - b^2(p-d^2) = p(d^2-b^2)\equiv 0 (\bmod p)$ .
Thus,
$(ad - bc)(ad+bc)\equiv 0 (\bmod p) \mbox{ thus }ad -bc = 0 \mbox{ or }ad+bc = p$ .

Our goal now is to show either $ad=bc \mbox{ or }ac=bd$ . Now $ad=bc$ follows at once from the equation $ad -bc =0$ . But if $ad+bc=p$ we will show this implies that $ac=bd$ .

Now if $p=ad+bc$ then $p^2 = (a^2+b^2)(c^2+d^2) = (ad+bc)^2+(ac-bd)^2 = p^2+(ac-bd)^2$ . Thus, $ac=bd$ .

Now we have established that $ad=bc \mbox{ or }ac=bd$ .

Say the first one is true $ad=bc$ . Then $a|bc$ . But since $\gcd(a,b)=1\implies a|c$ . Say therefore that $c=ak$ . Substitute $c=ak$ into $ad=bc$ we get $ad = abk \implies d=bk$ . But then $p = c^2+d^2 = (ak)^2+(bk)^2 = k^2(a^2+b^2)\implies k=1 \implies a=c \mbox{ and }b=d$ .

Now using a similar argument with $ac=bd$ we find that $a=d \mbox{ and }b=c$ .

I hope you like this proof. I really find it beautiful how divisibility arguments lead us to this conclusion.