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Are there any solutions for the diophantine equation $x^2+y^2=n$ ?

For $n \in \mathbb{P} \wedge n \equiv1\pmod4$ solutions are widely known.

Can we generalize a bit?

Bobson Dugnutt
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Martin Hopf
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    A lot. A positive integer $n$ is representable if and only if it has prime factorization of the form $2^e p_1^{a_1}\cdots p_k^{a_k}q_1^{2b_1}\cdots q_l^{2b_l}$, where the $p_i$ are primes congrient to $1$ modulo $4$,and the $q_i$ are primes congruent to $3$ modulo $4$. (It is possible to have $k=0$ and/or $l=0$.) – André Nicolas Mar 01 '16 at 17:26
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    See https://en.wikipedia.org/wiki/Fermat%27s_theorem_on_sums_of_two_squares and https://en.wikipedia.org/wiki/Brahmagupta-Fibonacci_identity. – lhf Mar 01 '16 at 17:30
  • Thank you! So we get answers only by a (partial) factorization of $n$ ? – Martin Hopf Mar 01 '16 at 17:34
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    See http://mathoverflow.net/questions/29644/enumerating-ways-to-decompose-an-integer-into-the-sum-of-two-squares. – lhf Mar 01 '16 at 17:41
  • Thank you! That is new for me. But I'm afraid we can't get rid of a factorization of $n$ at this point. – Martin Hopf Mar 01 '16 at 18:09

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