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I am working on a problem in number theory and would like to count all possible ways to partition an integer $n\geq 1$ into pairs $(k,m)$ of positive integers such that $n=2k^2+m^2$ and $n=4k^2+m^2$. Let's say $s(n)$ counts all pairs $(k,m)$ such that $n=2k^2+m^2$, while $t(n)$ counts all pairs $(k,m)$ such that $n=4k^2+m^2$. what are the generating functions of $s(n), t(n)$?

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    $\begingroup$ Is this a reasonable answer: $\sum_{n=1}^{\infty} s(n)x^n=\sum_{p=1}^{\infty} x^{p^2} \sum_{q=1}^{\infty} x^{2q^2}$? $\endgroup$
    – LeechLattice
    Commented Sep 2, 2021 at 22:56
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    $\begingroup$ This is not research-level math; further, it looks like homework. As such, it does not belong on mathoverflow, which is for research-level problems. It may be better suited for math.stackexchange; however, if you do choose to post there, you should give more context - especially by showing what you have tried to do and where you are stuck. See math.stackexchange.com/help/how-to-ask for more information. $\endgroup$
    – user44191
    Commented Sep 3, 2021 at 1:01
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    $\begingroup$ Uh, guys, this is a perfectly reasonable question. It could be assigned as homeowrk in an algebraic number theory course, but that doesn't mean that every mathematician knows how to do it, and it has much better answers than the one LeechLattice gives. $\endgroup$
    – David E Speyer
    Commented Sep 3, 2021 at 3:30
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    $\begingroup$ For $n = 4 k^2 + m^2$, look at the formula for the number of ways to write $n$ as $\ell^2+m^2$ mathworld.wolfram.com/SumofSquaresFunction.html . If $n \equiv 1 \bmod 4$, then half of these will have $\ell$ even and half will have $\ell$ odd; if $n \equiv 0 \bmod 4$, then all solutions have $\ell$ even; if $n \equiv 2 \bmod 4$ or $3 \bmod 4$, there are no solutions with $\ell$ even. $\endgroup$
    – David E Speyer
    Commented Sep 3, 2021 at 3:33
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    $\begingroup$ For $n = 2 k^2 + m^2$, there is a similar formula. Ignoring the condition that $k$ and $m$ be positive, we are counting elements in $\mathbb{Z}[\sqrt{-2}]$ of norm $n$. Since $\mathbb{Z}[\sqrt{-2}]$ is a PID and has unit group $\pm 1$, the number of elements of number $n$ is twice the number of ideals of index $n$. $\endgroup$
    – David E Speyer
    Commented Sep 3, 2021 at 3:35

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