$n!+1$ being a perfect square One observes that \begin{equation*} 4!+1 =25=5^{2},~5!+1=121=11^{2} \end{equation*} is a perfect square. Similarly for $n=7$ also we see that $n!+1$ is a perfect square. So one can ask th...--prophetes.ai

$n!+1$ being a perfect square One observes that \begin{equation} 4!+1 =25=5^{2},~5!+1=121=11^{2} \end{equation} is a perfect square. Similarly for $n=7$ also we see that $n!+1$ is a perfect square. So one can ask the truth of this question: * Is $n!+1$ a perfect square for infinitely many $n$? If yes, then how to prove.

This is Brocard's problem, and it is still open.

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