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I was doing some basic things on quadratic congruences. While doing so I encountered this.

Let $p$ be a prime and $r$ an integer such that $0\leq r<p$. Then I want a condition on $p$ (like $p\equiv a\mod b$, $a$ and $b$ are some positive integers) such that $$24 r +(k+1)=(6m+1)^2+k(6n+1)^2$$ has no solution for $m$ and $n$. Here, $k>1$ and $m$, $n$ are any integers.

Basically, I want $24r+(k+1)$ is a quadratic non residue modulo $p$ by putting some condition on $p$ from the above, for at least $k=2$ and $4$. Please help.

The conditions that support the above equality are

For $k=2$, $$24r+(k+1)\equiv 1 ~or~3 \mod 8,$$ and for $k=4$, $$24r+(k+1)\equiv 1 \mod 4.$$

Hirakjyoti Das
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  • What does $p$ have no relation to your condition. Should it be $\equiv\pmod p$ rather than $=$? – Thomas Andrews Sep 16 '19 at 18:39
  • The condtion I expect should be on the prime $p$. Actually, I was trying to find the condition on $p$ with the help of the last congruences. But, they came of no use. They are some results from solutions of $$N=x^2+2y^2$$ and $$N=x^2+4y^2.$$ – Hirakjyoti Das Sep 16 '19 at 18:43

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