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

Question

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? $$p = (a-b)(a+b)$$

Answer 1

If $(a+b)(a-b) = p$ than either $(a+b) = 1$ or $(a-b) = 1$. Say $(a-b) = 1$, so $b = a-1$. Then we have $2a-1=p$, which has infinitely many solutions, namely $a=\frac{p+1}{2}$ for every prime p besides $2$.

Answer 2

Suppose there exists integers $a,b$ and prime $p$ such that

$$p = (a-b)(a+b)$$

This implies that $(a-b)$ and $(a+b)$ are factors of $p$. Since $p$ is prime, then one of $(a-b)$ and one $(a+b)$ must be $1$ and the other $p$, otherwise $p$ is not prime. Thus, one of the two systems must be the case:

$$\left\{\begin{matrix} a+b = 1 \\ a-b = p \end{matrix}\right. \;\;\; \text{or, alternatively} \;\;\; \left\{\begin{matrix} a-b = 1 \\ a+b = p \end{matrix}\right.$$

Solve the system on the left: you'll find $a = (p+1)/2, b = (1-p)/2$.

Solve the system on the right: you'll find $a = (p+1)/2, b = (p-1)/2$.

Thus, for any given $p$, there are only two solutions for $a,b$ in integers. provided $p>2$. If $p=2$, then neither $a$ or $b$ are integers.

If you mean solutions in terms of all three variables - $a,b,p$ - then there are infinitely many since you can relate $a,b$ in terms of any given $p>2$.

Answer 3

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 $a=-\frac{p+1}2; b = \frac {p-1}2 = -1-a$