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June 25th, 2015, 05:53 PM  #1 
Newbie Joined: Dec 2014 From: Singapore Posts: 23 Thanks: 0  Expressing integers as sum of two squares
I've been thinking about this problem for some time and need help with it: Let $n$ be a natural number, and $(a_i, b_i)$ be the solutions to $x^2+y^2=n^2$,$0<a_i<b_i$. If $i$ has the range $1\leq i\leq k$, then: Can $k$ be larger than a given natural number for large enough $n$? If so, does there exist an $n$ for every $k$? Then, is there a expression for $n$ in $k$? 
June 25th, 2015, 08:30 PM  #2 
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 
Are $a_i$ and $b_i$ real or natural numbers?

June 26th, 2015, 10:27 PM  #3 
Newbie Joined: Dec 2014 From: Singapore Posts: 23 Thanks: 0 
Natural numbers

June 27th, 2015, 04:11 PM  #4  
Global Moderator Joined: May 2007 Posts: 6,821 Thanks: 722  Quote:
You are looking at Pythagorean triples. The above is a list of them. For most n, k = 0. I didn't scan it thoroughly, but the largest k I saw was 4.  
June 27th, 2015, 07:28 PM  #5 
Newbie Joined: Dec 2014 From: Singapore Posts: 23 Thanks: 0 
I am looking for the number of Pythagorean triples that have a given hypotenuse. In other words, k is the number of lattice points the equation x^2+y^2=n^2 passes through in the first quadrant above the line x=y. Other points can be deduced by symmetry.

June 28th, 2015, 05:53 AM  #6  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
I think that the answer is yes. Certainly f(n) takes on arbitrarily large values, but probably it doesn't leave any gaps. The first values, if I'm not mistaken, are f(5) = 1 f(65) = 2 f(325) = 3 f(1105) = 4 f(8125) = 5 f(5525) = 6 f(105625) = 7 f(27625) = 8 f(71825) = 9 f(138125) = 10 f(?) = 11 f(160225) = 12 f(1221025) = 13 f(3453125) = 14 f(1795625) = 15 f(801125) = 16  

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