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What numbers $n$ can be written as a sum of odd squares:

$$a^2 +b^2 = n$$ Where $a$ and $b$ are odd.

It is easy to see that all such numbers must be of form $2(4k+1)$ and if $4k+1$ is prime it can be written as a sum of $2$ squares. Beyond that, I am stumped.

mtheorylord
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    @John: note that this question asks for numbers which are the sums of two odd squares. I'm not very sure and this might turn out to be equivalent trivially to the usual case, but regardless I don't think this qualifies as a duplicate. – YiFan Tey Nov 08 '19 at 00:09
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    @YiFan Thanks for your feedback. However, note the sums of two odd squares gives $n \equiv 2 \pmod 4$, as the OP has stated. Since the sum of $2$ even, or one even & one odd, squares will not give this, handling of such $n$ will only result in odd squares, as requested. The linked to question handles all possible numbers, including this case, with not too many complications to account for this simpler one, so I still consider it to effectively be a duplicate. – John Omielan Nov 08 '19 at 00:13
  • @John: I see your point, and that observation you just made may have been trivially obvious to you, but it might not be for others. Evidently it was not for the OP. – YiFan Tey Nov 08 '19 at 00:15

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Any product $2PQ$ where $P$ is a product of primes of the form $4k+1$ and $Q$ is an odd square will fit the bill.

You can prove the result for the odd primes in the usual way and then multiply by $1^2+1^2$, again in the usual way, to obtain a sum of odd squares.