November 24th, 2017, 02:28 AM  #1 
Newbie Joined: Oct 2017 From: sweden Posts: 14 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 $<a_0,a_1,....., a_n>$. 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 $h_ipuk_i$ 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. 
December 1st, 2017, 04:04 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,875 Thanks: 766 
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.


Tags 
proving, y2p 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Proving.  jiasyuen  Algebra  4  December 24th, 2014 08:02 PM 
proving big O  nealelliott  Computer Science  1  March 6th, 2014 06:44 AM 
proving 0=0  chris123  Algebra  24  February 27th, 2012 04:04 PM 
Proving something is a subspace by proving it is a span.  suomik1988  Linear Algebra  1  April 29th, 2011 10:57 PM 
proving log(a)  ^e^  Real Analysis  6  March 1st, 2007 09:16 PM 