I was stuck at this question:

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> Suppose $a^2+b^2=c^2$ for $a,b,c \in \mathbb Z$, and neither $a$ nor $b$ is a multiple of 7. Show that $a^2-b^2$ is a multiple of 7
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I tried to write $b^2$ as $c^2-a^2$ then get $a^2-b^2=2a^2-c^2$. But this does not seem to generate the solution.

How to solve problems like this, am I missing some theorems concerning Pythagoras numbers?