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 November 24th, 2017, 01:28 AM #1 Newbie   Joined: Oct 2017 From: sweden Posts: 24 Thanks: 0 proving x^2+y^2=p There seems to be many variations of the proof to this problem, but how would one prove this using continued fractions? The problem stated in my book is proved using continued fractions, but I don't get it: It says when $p = 1\bmod 4$, $x^2+y^2 = p$. But it said we should suppose that there is some $u^2=-1 \bmod p$, and $u/p$ is the continued fraction . It then says we should first show that this inequality holds $|h_i/k_i -u/p|<\frac{1}{k_i\sqrt{p}}$, which implies \$0
 December 1st, 2017, 03:04 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Please state exactly what it is that you are trying to prove! It is probably "there exist integers x and y such that ..." but I am not sure.

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