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 Factorials and Powers (Posted on 2009-04-22)
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 = XY

 No Solution Yet Submitted by K Sengupta Rating: 4.0000 (2 votes)

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 Very little progress to report | Comment 6 of 8 |

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^(Y-1) + X^(Y-2) + ..... + X + 1] gives:
(X - 2)! = X^(Y-1) + X^(Y-2) + ..... + 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 2009-04-24 00:15:13

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