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June 25th, 2015, 05:53 PM   #1
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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 $k$?
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June 25th, 2015, 08:30 PM   #2
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Are $a_i$ and $b_i$ real or natural numbers?
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June 26th, 2015, 10:27 PM   #3
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Natural numbers
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June 27th, 2015, 04:11 PM   #4
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Quote:
Originally Posted by Chris A View Post
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 $k$?
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.
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June 27th, 2015, 07:28 PM   #5
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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.
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June 28th, 2015, 05:53 AM   #6
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Quote:
Originally Posted by Chris A View Post
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 $k$?
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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