<s>[INST] Proving prime divisibility relation between $a^2-a+3$ and $b^2-b+25$. Let $p$ be a given prime number. Prove that there exists an integer $a$ such that $p|a^2-a+3$ if and only if there exists an integer $b$ such that $p|b^2-b+25$. 
I've managed to prove that if $p|a^2-a+3$ for some $a$, then $p|b^2-b+25$ for some $p$ like so:
Working mod $p$, we have $a^2-a+3=0$. Hence $9a^2-9a+27=0$, which implies $9a^2-6a+1=3a-26$. Rearranging this, we find that $$(3a-1)^2=(3a-1)-25$$ so simply taking $b=3a-1$ proves this part.
I had the idea of multiplying everything by $9$ because $3\times 9\approx 27$, and because $9$ is a square. However, I can't find a similar relation for turning $b^2-b+25$ into $a^2-a+3$.
According to the person I found this question from, this question is from a very old TST (I think it was from China, but we're not sure).
 [/INST] $a^2-a+3\equiv0\pmod p$
$\iff(2a-1)^2\equiv-11$   for odd $p$
$4(b^2-b+25)=(2b-1)^2+99$
So, we need $(2b-1)^2\equiv-99\pmod p$
$\implies(2b-1)^2\equiv3^2(2a-1)^2\pmod p$
$\iff2b-1\equiv\pm3(2a-1)$
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