Brocard's Problem. Is $n!+1={m}^{2}$ ?

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?

Note: this problem is also known as Brocard's Problem and it's still open in list of unsolved math problems.

#Algebra #NumberTheory #Factorial #ComputerScience #UnsolvedMysteries

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

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Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Brocard's Problem. Is n!+1=m2n!+1={m}^{2}n!+1=m2?

Note: this problem is also known as Brocard's Problem and it's still open in list of unsolved math problems.

Comments

Brocard's Problem. Is $n!+1={m}^{2}$ ?