Find $p$ : $p\mid k^2-5$

Question

Which odd primes can divide the integers of the form $k^2-5$ ?

My try: odd prime means $p>2$ and if $$p\mid k^2-5$$ then $$k^2\equiv5 \mod p$$ hence 5 is quadratic residue mod p

Answer 1

By quadratic reciprocity $$\left(\frac5p\right)\left(\frac p5\right)=(-1)^{\frac{p-1}2\frac{5-1}2}=(-1)^{p-1}=1$$ Since $\left(\frac 5p\right)=1$, so is $\left(\frac p5\right)$, i.e. $n^2\equiv p\bmod5$ for some $n$. For this $n$ to exist, either $p=5$ and thus $n=0$, or $p=\pm1\bmod5$ since the nonzero squares in $\mathbb F_5$ are $\pm1$. Since $p$ is odd, we can refine the latter statement to $p=\pm1\bmod10$.

Thus, the odd primes dividing integers of the form $k^2-5$ are $5$ and those whose decimal representations end in $1$ and $9$.