Difference of square makes prime

If $p$ is a prime number, it's factors are $p$ and $1$

Then we can write, $m^2-n^2=p\\(m+n)(m-n)=p\times 1$

Afterthat, it must be true that $m+n=p\\m-n=1$

If we solve them, we get $m=\frac{p+1}{2}\\n=\frac{p-1}{2}$

For $p>2$ all prime number must be odd number. So $p+1,p-1$ must be divisible by $2$ and we always get $m,n$ as integer.

So the statement is $\color{#20A900}True$