# Nonzero quadratic residues modulo 101

How many Nonzero quadratic residues are there modulo prime 101

I am lost where to start to my knowledge there is no formula for number of quadratic residues a prime has It will be too much to start checking all numbers from 2 to 100 if they are a quadratic residue or not

Any help appreciated and please also guide me if there is a way to do this for any prime p Thanks

Expanding David's answer, for any $p>2$ the map $\phi:\mathbb{F}_p^*\to\mathbb{F}_p^*$ defined by $\phi(x)=x^2$ sends $y$ and $-y$ into the same element, hence the number of quadratic residues is $\leq\frac{p-1}{2}$. On the other hand, $\mathbb{F}_p^*$ is a cyclic group generated by $g$ with $o(g)=p-1$, hence all the "even powers" of $g$ are quadratic residues, giving that the number of quadratic residues is $\geq\frac{p-1}{2}$.
Modulo a prime (except for $2$), exactly half the non-zero residues are quadratic residues.
Note that if $r\equiv x^2\equiv y^2 \mod p$ then $x^2-y^2=(x+y)(x-y)\equiv 0 \mod p$
Since $p$ is prime, one of the factors $x+y, x-y$ must be divisible by $p$, and this is equivalent to $x\equiv \pm y \mod p$.
Apart from the residue zero, and the case $p=2$ where we have $1=-1$, this gives precisely two square roots for each quadratic residue. The number of square roots is $p-1$ (just square them all), so the number of quadratic residues is $\cfrac {p-1}2$