2023/008) find integer n such that $(n-1)! + 1 = n^2$

We have  $(n-1)! = n^2-1 = (n+1)(n-1)$

So n-1 =0 which gives LHS = 2 RHS = 1 which is contradiction 

Or $(n-2)! = n + 1$

Put n- 2 = k giving $k! =k + 3$

As LHS is divisible by so is RHS so k is a factor of 3 k = 1 or 3

k =1 gives n = 3 which is not a solution as it does not satisfy the criteria

k =3 gives n= 5 and as $4! + 1 = 25 = 5^2$ so n = 5 is a solution