Can $b+c$ in the pythagorean triplets $(a, b, c)$ be a prime number?

Question

If $a^2 + b^2 = c^2$, can $b + c$ be a prime number?

Answer 1

If the Pythagorean Triple $(a, b, c)$ is not primitive, then $b+c$ is trivially composite. Thus, let it be primitive. Then $$a=m^2-n^2$$ $$b=2mn$$ $$c=m^2+n^2$$ where $b$ and $a$ are interchangeable. Then we are looking at the sum $b+c=2m^2$ or $(m+n)^2$

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Looking back at this answer, the only possibility for $b+c$ to be prime is when $m=1, b=1-n^2$, and thus $b+c=2$.

Answer 2

Hint: $a^2+b^2=c^2\implies b^2-c^2=-a^2\implies (b+c)(b-c)=-a^2 \implies b+c=\frac{a^2}{c-b}$ $ \implies b+c=a^2 \times \frac{1}{c-b}$