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If $a=2$ and $b=3$, then $a^2-1$ is an integer multiple of $b$. Is there any other pair of primes $a$ and $b$ that satisfy this relationship? I don't think so, but can't figure out why not.

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There are plenty of them, infinite of them to be exact. Set $a= \text{any prime} =p$ and $b$ to any prime divisor of $p^2-1$.

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  • $\begingroup$ True, thanks! What if $a<b$? $\endgroup$ – user101600 Dec 8 '13 at 21:58

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