Brocard's Problem

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Unsolved Problem

For which pairs of (strictly) positive integers $\tuple {m, n}$ do the following hold:

$n! + 1 = m^2$

The only known pairs are:

\(\text {(1)}: \quad\) \(\, \displaystyle \tuple {5, 4}: \, \) \(\displaystyle 4! + 1\) \(=\) \(\displaystyle 24 + 1 = 25 = 5^2\)
\(\text {(2)}: \quad\) \(\, \displaystyle \tuple {11, 5}: \, \) \(\displaystyle 5! + 1\) \(=\) \(\displaystyle 120 + 1 = 121 = 11^2\)
\(\text {(3)}: \quad\) \(\, \displaystyle \tuple {71, 7}: \, \) \(\displaystyle 7! + 1\) \(=\) \(\displaystyle 5040 + 1 = 5041 = 71^2\)


Also see


Source of Name

This entry was named for Pierre René Jean Baptiste Henri Brocard.


Sources