Determine all possible pair(s) (X, Y), with X being a prime and Y being a positive integer, that satisfy this equation:

(X-1)! + 1 = X^Y

we want

(x-1)!=(x^y)-1

so for y=1 we have

(x-1)!=x-1

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

(x-1)!=x^2-1=(x-1)(x+1) dividing by x-1 we get

(x-2)!=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 2009-04-22 14:19:02