Is there any $n$ such that the equation $n = x^2 + y^2$ ($n, x, y$ belonging to the set of natural numbers) has more of three solutions?
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If you mean ordered solutions, $0 < x < y,$ take $n = 5 \cdot 13 \cdot 17 \cdot 29 $ – Will Jagy Aug 20 '13 at 00:22
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Depends on how you are counting - are $(x,y)=(5,0)$ and $(x,y)=(0,5)$ different answers for $n=25$ The answer, however is use, for "more than $k$" for any $k$. In particular, there are cases where there are more than $3$ solutions. – Thomas Andrews Aug 20 '13 at 00:22
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@WillJagy Do you really need three primes? I think $5\cdot 13\cdot 17$ is enought. – Thomas Andrews Aug 20 '13 at 00:23
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@Thomas, could be. I don't remember whether you double with each prime or just add one. – Will Jagy Aug 20 '13 at 00:24
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Definitely, multiply :) @WillJagy – Thomas Andrews Aug 20 '13 at 00:26
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I think I read that there are $n$ having more than $k$ distinct solutions, for any positive $k$. (This seems clear by multiplication of solutions, but maybe some care to make sure they're all different solutions.) – coffeemath Aug 20 '13 at 00:27
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Yes, if $p_i\equiv 1\pmod 4$ are distinct primes then $n=p_1p_2\dots p_k$ has $2^{k-1}$ solutions (if you count switching $x,y$ as the same answer.) If $C_n$ is the count for $n$ then if $n,m$ are relatively prime then $C_{nm}=C_{n}C_{m}$ - at least, if $C_n$ allows $x=0$. – Thomas Andrews Aug 20 '13 at 00:30
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5So: $n=5\cdot 13\cdot 17 = 4^2+33^2 = 9^2+32^2 = 12^2+31^2 = 23^2 + 24^2$. – Thomas Andrews Aug 20 '13 at 00:32
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@ThomasAndrews, yes. – Will Jagy Aug 20 '13 at 00:34
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See here, here and here for some of the local hits to a suitable search. Degree of relevance varies. – Jyrki Lahtonen Aug 20 '13 at 12:13
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All natural numbers $n=p_1^{e_1}p_2^{e_2}\cdots p_r^{e_r}$ with primes $p_i\equiv 1(4)$ for all $i$ and $B:=(e_1+1)(e_2+1)\cdots (e_r+1)\ge 8$ have at least $4$ different representations as sum of $2$ squares, in fact $B/2$ different representations, if $B$ is even, or $(B-1)/2$ if $B$ is odd. Of course, there are other $n$ with that property as well. So far, this was more or less said already in the comments. However, if you are interested on how to obtain these representations computationally, then you might want to see this discussion here: https://mathoverflow.net/questions/29644/enumerating-ways-to-decompose-an-integer-into-the-sum-of-two-squares.
Dietrich Burde
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