Given a prime $p$. Prove that there exist $\alpha$ such that $p|\alpha(\alpha-1)+3$, if and only if there exist $\beta$ such that $p|\beta(\beta-1)+25$.
My solution:
Using quadratic residuu we have that $$\alpha=\frac{1\pm \sqrt{-11}}{2}$$ $$\beta=\frac{1\pm \sqrt{-99}}{2}=\frac{1\pm 3\sqrt{-11}}{2}$$ Since $p$ must be an odd prime, $\frac{1}{2}$ has inverse mudulo $p$. Supose that there exists $\alpha$ then $\sqrt{-11}$ exists modulo $p$. Hence $\beta$ also exists. The reciprocal is the same.
Could anyone please check this solution? And does anyone has another solution?
Thanks!
For $p=2$, or any prime where $3=25$, the conditions are identical.
For odd primes,
$x^2 - x = a$ is solvable if and only if $4a+1$ is a perfect square.
The statement is now that $-11$ and $-99$ are both squares or both nonsquares. For $p \neq 3$ that is clear and for $3$ one can calculate.
So there is some special verification at $p=2$ (where completing the square does not work) and $p=3$ (where multiplication by 9 can change a nonsquare into a square) and the argument using quadratic equations or completing the square handles all other values of $p$.