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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 $x^2+y^2=n^2$,$0  June 25th, 2015, 08:30 PM #2 Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 Are$a_i$and$b_i$real or natural numbers?  June 26th, 2015, 10:27 PM #3 Newbie Joined: Dec 2014 From: Singapore Posts: 23 Thanks: 0 Natural numbers June 27th, 2015, 04:11 PM #4 Global Moderator Joined: May 2007 Posts: 6,821 Thanks: 722 Quote:  Originally Posted by Chris A I've been thinking about this problem for some time and need help with it: Let$n$be a natural number, and$(a_i, b_i)$be the solutions to$x^2+y^2=n^2$,$0
Integer Lists: Pythagorean Triples

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
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Quote:
 Originally Posted by Chris A 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
Let f(n) be the number of representations of n as a sum of two distinct nonzero squares. Then the question, as I understand it, is: For any integer k >= 0, is there some n such that f(n) = k?

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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