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November 24th, 2017, 02:28 AM   #1
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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 $<a_0,a_1,....., a_n>$. 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 $0<x^2+y^2<2p$ when we substitute the values of $h_i$ with $x$ and $h_ip-uk_i$ with $y$, (note $i$ is the largest integer such that $k_i \le \sqrt{p}$). Then we have to deduce that indeed $x^2+y^2 = p$.

How does this approach to proving $x^2+y^2 = p$ when $p = 1\bmod 4$ work exactly?

Last edited by skipjack; December 1st, 2017 at 04:51 AM.
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December 1st, 2017, 04:04 AM   #2
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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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