Finding number of distinct primes: a paradoxical solution

Question

Let $a$, $b$, be primes such that the difference between their squares is prime. How many distinct values of $b$ are there?

This problem basically asks for distinct solutions to:

$a^2 - b^2 = p$, where $p$ is prime

or equivalently:

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

Since a prime has only two factors, $1$ and itself, $a - b$ is $1$

($a + b$ can't be $1$ since no primes satisfy that equation)

So, $a$ and $b$ are consecutive, there's only one pair of consecutive primes, so one solution, easy.

But, if we go back to $(a + b)(a - b) = p,$

since $a - b$ is $1,$ $a + b$ is $p$

or $b + b + 1 = 2b + 1$ is prime

Here, there are actually infinitely many values of b which would yield a prime(every prime is of the form $2q + 1$).

Can someone point out the error in this? There has to be something I'm doing wrong here.

Answer 1

There’s nothing paradoxical here. The only reason that you know that $a+b=2b+1$ is that you’ve already deduced that $a-b=1$. Thus, you can’t conclude that every prime of the form $2b+1$ yields a solution: it’s only the ones for which $b$ and $b-1$ are both prime, and the only one of those is $b=3$.

Answer 2

Yes, $2b+1$ is prime, but then it must also satisfy $a-b=1$. And since $a$ and $b$ are primes anyway, there can be only one solution to them. The reasoning on $2b+1$ needs to take into account conditions previously established – and said reasoning adds nothing to the solution.