Can you show that there exists an integer \(n\) other than 4, 5 and 7 such that \(n!+1={m}^{2}\). In other words, \(n! + 1\) is a perfect square?

Can you show that there exists an integer \(n\) other than 4, 5 and 7 such that \(n!+1={m}^{2}\). In other words, \(n! + 1\) is a perfect square?

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