Determine all possible pair(s) (X, Y), with X being a prime and Y being a positive integer, that satisfy this equation:
(X1)! + 1 = X^{Y}
I agree with the solutions already found: (2, 1), (3, 1), (5, 2), but have almost given up hope of finding others or disproving their existence. Is anyone else still trying?
I know that Wilson's theorem states that the congruency:
(X  1)! + 1 = 0 (mod X) is true iff X is prime (amazing). Since X is a factor of our RHS, X^Y, it follows that any integer solutions for X must be prime  so the guidance in our question, to look for prime values, was not strictly necessary  but was kind.
I've also realised that, for X>1, the rearrangement:
(X  1)! = X^Y  1 = (X  1)[X^(Y1) + X^(Y2) + ..... + X + 1] gives:
(X  2)! = X^(Y1) + X^(Y2) + ..... + X + 1
Now, for X>3, the LHS is even and, since all terms on the RHS are odd, there must be an even number of terms; so Y is even (Y = 2y say).
Our equation now becomes: (X  1)! + 1 = (X²)^y
Mid 20th Century number theorists (Hardy et al..) were writing that the congruency: (X  1)! + 1 = 0 (mod X²) had solutions X = 5, 13, 563,... but had found no others below 200000. Since X² is now a factor of our RHS, it follows that our values of X must be a subset of these. However, of these three, only the '5' satisfies the greater constraints imposed by our RHS, and we already knew of that solution  so no joy there.
At that time they also wrote that no general theorem had been found to deal with this type of congruency  so please can anyone bring me up to date..?

Posted by Harry
on 20090424 00:15:13 