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By the theorem of Fermat, odd prime $p$ can be represented as $p=x^2+y^2$($x,y$ : integer, $0\leq x\leq y$) if and only if $p \equiv 1$ ($\text{mod } 4$). Moreover, such decomposition is unique.

I would like to know that if the mapping $\{p|p:\text{prime such that }p\equiv 1 (\text{mod } 4)\} \ni p=a^2+b^2 \mapsto a+b\in \{2n+1|n\geq 1\}$ is surjective.

My research.

My first idea was to use the explicit formula of this decomposition. This thread suggests

$a=\frac12\binom{2k}k\pmod p$, $b=(2k)!a\pmod p$ ($|a|,|b|<p/2$).

is the formula. So $a+b = \frac{1}{2}\binom{2k}k+(2k)!\frac{1}{2}\binom{2k}k \pmod p$?

Though how can we prove that it is surjective?

Kaira
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    Another way of wording this is that for any positive integer $n$, we want there to exist a positive integer $a<2n+1$ so that $a^2+(2n+1-a)^2$ is prime. In general determining whether quadratics generate primes is difficult, so I’d be pleasantly surprised if there were an elementary solution. – Eric Jun 21 '21 at 04:19

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