$\forall p, b, c \in \mathbb{Z}^+, (\text{Prime}(p) \wedge p^2 + b^2 = c^2) \Rightarrow p^2 = c+b$

Question

How do I prove that $\forall p, b, c \in \mathbb{Z}^+, (\text{Prime}(p) \wedge p^2 + b^2 = c^2) \Rightarrow p^2 = c+b$ ? I know this much; $$p^2 + b^2 = c^2\\ p^2= c^2 - b^2\\ p^2=(c-b)(c+b) $$ I know the next part has something to do with the fact that every number that is not prime can be expressed as a product of primes that is unique(up to the orders of the factor?). And using that, somehow, $c-b=1$ and, therefore, $p^2 = c+b$. If someone could please offer me a rigorous explanation for this it would be appreciated.

Answer 1

If $p^2=(c-b)(c+b)$ and $c-b\neq 1$, then $c-b=c+b=p$ because $p $ is prime. This implies $b=0$, a contradiction which proves $c-b=1$ and $c+b=p^2 $.

Answer 2

Part 1. It's easy to see that $p > c-b$, otherwise, if we assume $p \leq c-b \Rightarrow p+2b\leq c+b$ and $p(p+2b)\leq(c-b)(c+b)=p^2$, which is possible only when $b=0$.

Part 2. From $p^2=(c-b)(c+b) \Rightarrow p \mid (c-b)(c+b)$. Applying Euclid's lemma we have $p \mid (c-b)$ or $p \mid (c+b)$. But, due to part 1, $p \mid (c-b)$ doesn't make sense (as well $c-b=0$, leading to $p=0$), thus $p \mid (c+b)$.

Part 3. Applying Euclid's lemma and Part 2 again, $p^2 \mid (c+b)$ or $\exists q\in\mathbb{Z^+}: c+b=p^2q$ or $p^2=(c-b)(c+b)=(c-b)p^2q \Rightarrow 1=q(c-b)$ or $c-b \mid 1$ which is possible only when $c-b=1$ and $q=1$, thus $p^2=c+b$.