Funny Prime

The well-known factorization $a^2 - b^2 = (a+b) \times (a-b)$ turns the equation into $(a+b) \times (a-b) = p$ .

We know that the condition for p to be prime is that his only divisors are 1 and himself, meaning that $a+b$ and $a-b$ are either $1$ or $p$ .

$a+b = 1$ is absurd since $a$ and $b$ are positive integers and that would mean that $a = 0$ and $b=1$ or the other way around, and in this case, $a^2-b^2$ would not be a prime.

So we know $a+b=p$ and $a-b=1$ . Putting $a = b + 1$ back into the equation gives $(b+1)^2-b^2=p$ , that's to say $2b+1 = p$ so $b=\frac{p-1}{2}$ and then we deduce $a=\frac{p+1}{2}$ .

a^2- b^2 = (a+b)(a-b)= p. So p has 2 factors one is a+b and other is a-b. For p to be prime the factors must be 1&p itself.

So a+b=p and a-b=1

So p= 2b+1 b = (p-1)/2 And a = (p+1)/2