# $n^2 \equiv (-5) \mod p$

I need to determine all primes p for which -5 is a quadratic residue modulo p. The original question is

Find the odd primes which divide integers of the form $$n^2+5$$ ?

My try:

what I obtain is ... $$n^2 \equiv (-5) \mod p$$ I know how to solve $$n^2 \equiv 5 \mod p$$ what is the difference if it is replaced by -5 . Does it mean... every prime of the form $$p=\pm 1 , \pm3 \mod 10 ??$$

• Do you know about the Legendre symbol and specifically its multiplicative properties? May 31, 2019 at 8:13
• That would be every odd prime except $5$, which is not right. First note that $5$ must be included (eg the square $0^2=0\equiv -5 \bmod 5$ - it is easy to miss special cases. It would be helpful if you could show some workings - at the moment it is hard to see what you may have done - but the quadratic character of $-1$ modulo $p$ depends on $p\bmod 4$ May 31, 2019 at 8:13

We have the Legendre symbol $$\left(\frac{-5}p\right)=\left(\frac5p\right)\left(\frac{-1}p\right)=+1$$

There are two main cases:

• If both Legendre symbols are $$+1$$, then (as already shown) either $$p=5$$ (the exceptional case) or $$p\equiv\pm1\bmod5$$. $$\left(\frac{-1}p\right)=+1$$ precisely when $$p\equiv1\bmod4$$. Combining these two congruences yields $$p\equiv1,9\bmod20$$.
• If both Legendre symbols are $$-1$$, solving each symbol yields $$p\equiv\pm2\bmod5$$ and $$p\equiv3\bmod4$$, or $$p\equiv3,7\bmod20$$.

Therefore, the odd primes that are factors of $$n^2+5$$ are $$5$$ and those congruent to $$1,3,7,9\bmod20$$.

The possible prime factors upto $$5$$ are $$2,3,5$$ which can be found out by inspection.

For primes $$p>5$$ , we have to distinguish between two cases :

$$(1)$$ $$p=4k+1$$. In this case, $$-1$$ is a quadratic residue and hence also $$5$$. So, the possible forms are $$20k+1$$ and $$20k+9$$

$$(2)$$ $$p=4k+3$$. In this case, $$-1$$ is not a quadratic residue, so neither is $$5$$. So, the possible forms are $$20k+3$$ and $$20k+7$$