For integers $a,b$ and prime $p$, are there infinitely many solutions to $p = (a-b)(a+b)$? I was thinking on this problem: suppose, for a prime $p$ and integers $a,b$, we have
$$p = (a-b)(a+b)$$
This implies immediately that $p \mid (a-b)$ or $p \mid (a+b)$. Then, since $p$ is prime and thus its only factors are $1$ and itself, this means $a+b = p$ and $a-b = 1$.
From that, $a=b+1.$ It's easy to see that, say, if $a$ ends in $3$ and $b$ ends in $2$, or if $a$ ends in $8$ and $b$ ends in $7$, that do not necessarily satisfy the original equation, just $a=b+1$. They don't necessarily satisfy $a+b=p$, however.
So my question is:

For integers $a,b$ and prime $p$, are there infinitely many solutions to the equation below?
  $$p = (a-b)(a+b)$$


Of course not!
If $p$ is prime so its only factors are $\pm 1$ and $\pm p$ so so if $p = (a-b)(a+b)$ then  $a+b$ and $a-b$ can only be $\pm 1$ and/or $\pm p$. 
So either $a-b = 1$ and $a+b = p$ so $a=\frac {p+1}2; b = \frac {p-1}2 = a-1$.
Or $a-b = p$ and $a + b = 1$ so $a= \frac {p+1}2; b = -\frac {p-1}2= 1-a$
Or $a-b = - 1$ and $a+b = -p$ so $a =-\frac {p+1}2; b=-\frac {p-1}2=a+1$
or $a-b = -p$ and $a + b = -1$ so $a=-\frac{p+1}2; b = \frac {p-1}2 = -1-a$
