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

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  • $\begingroup$ If $a^2+b^2=c^2$ is primitive, then $b+c=(2x-1)^2,x\in\mathbb{N}$ $\endgroup$
    – poetasis
    Mar 16, 2020 at 18:23

2 Answers 2

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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 $(m+n)^2$


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

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  • $\begingroup$ Thanks for your help. Appreciate it a lot. $\endgroup$
    – T.C.
    Jun 12, 2016 at 7:28
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    $\begingroup$ No problem. If my answer was helpful, you should check the checkmark under the arrows for my comment. It gives me more reputation $\endgroup$
    – JasonM
    Jun 12, 2016 at 7:30
  • $\begingroup$ @almagest Actually it wasn't correct. I just corrected it. Sorry everyone. $\endgroup$
    – JasonM
    Jun 12, 2016 at 23:10
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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}$

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  • $\begingroup$ A lemma that will solve the problem when coupled with the hint: if $p$ is prime and $p$ divides $a^2$, then $p$, being a divisor of $a$, is less than or equal to $a$. $\endgroup$
    – Aaron
    Jun 12, 2016 at 8:12

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