2007 JBMO Problems/Problem 4

Problem 4

Prove that if $p$ is a prime number, then $7p+3^{p}-4$ is not a perfect square.


By Fermat's Little Theorem, $7p+3^p-4\equiv3-4\equiv-1\mod p$. By quadratic residues, this is true if and only if $p\equiv1\mod4$, except for $p=2$ (which doesn't work). Then, $7p+3^p-4\equiv3+3-4=2\mod4$, but this implies $v_2(7p+3^{p}-4)$ is odd, so $7p+3^{p}-4$ cannot be a perfect square.

See also

2007 JBMO (ProblemsResources)
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Problem 3
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