If is a natural number such that it follows the constraints (n-1)! + 1 is a multiple of and is also a perfect square, what is the value of ?
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According to Wilson's theorem, (n-1)! + 1 is divisible by n if and only if n is prime. Hence, n is prime.
Until now, there are only 3 known values of (n-1)! + 1, which are perfect squares:
4! + 1 = 25
5! + 1 = 121
7! + 1 = 5041
Therefore, from these values, the values of n in (n-1)! + 1 are 5, 6, & 8. Since n is prime, only number 5 works.
As a result, n = 5.