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}
we want
(x1)!=(x^y)1
so for y=1 we have
(x1)!=x1
or n!=n and this is true only for n=1 and n=2 thus we have
(2,1) and (3,1) as solutions
for y=2 we have
(x1)!=x^21=(x1)(x+1) dividing by x1 we get
(x2)!=x+1 and this is only true for x=5 thus we have
(5,2) as another solution
I am working on a proof that there are no further solutions for y>2 but do not have time right now to complete it.

Posted by Daniel
on 20090422 14:19:02 