If a prime number $p$ that isn't a Gaussian prime number,and can be decomposed into the sum of a square number $m^2$ and a prime number $q$,whether the prime number $q$ isn't a Gaussian prime number？For example,$p=113$,$m=4$,$q=97$;$p=113$,$m=10$,$q=13$.
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1What's your question, then? – weareallin Mar 09 '19 at 17:18
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For those unfamiliar with the term, a "Gaussian prime" is a prime in the ring of Gaussian integers, $\mathbb{Z}[i]$. Since $2$ or any prime congruent to $1\pmod 4$ factors in the Gaussian integers, the only primes in $\mathbb{Z}$ which are Gaussian primes are those which are congruent to $3 \pmod 4$. (See, for example, this question.)
So the question becomes this: If $p \not\equiv 3 \pmod 4$ is a prime and $p = m^2 + q$ with $q$ prime, then is $q \not\equiv 3 \pmod 4$?
The answer is yes. If $p = 2$, the only possibility is $m=0$, $q=2$. Otherwise, $p\equiv 1 \pmod 4$. We cannot have $m$ odd, as that would imply $m^2 \equiv 1 \pmod 4$, so $q$ would be a multiple of $4$. If $m$ is even, then $m^2\equiv 0\pmod 4$, so $q\equiv 1\pmod 4$.
FredH
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