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 Factorial-Square Sum (Posted on 2014-03-23)
Determine all possible triplets (X,Y,Z) of positive integers that satisfy:

X!*Y! = X! + Y! + Z2

Prove that there are no others.

 No Solution Yet Submitted by K Sengupta No Rating

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 Some analytical conclusions Comment 7 of 7 |
Without loss of generality, let X <= Y.  Then X! divides the left hand side, and also X! and Y!, so it must divide Z^2.

Dividing by X! gives
Y! = 1 + Y!/X! + Z^2/X!

let p be the smallest (first) prime > X/2

Consider each of the sides of the equation, mod p.

Y! (mod p) = 0
A little less obviously, Z^2/X! (mod p) = 0.
This is because z^2 must be a multiple of p^2, while x! is a multiple of p (but not p^2)

Therefore Y!/X! cannot be a multiple of p, because the equation mod p become 0 = 1 + 0 + 0.  Therefore, Y < 2p.

Also, Y cannot equal X, because the equation mod p becomes 0 = 1 + 1 + 0.
This is possible if p = 2, but this leads to the solution (2,2,0) which is not allowed because Z is not positive.

Thus, Y must > X but very close to X.

For instance, if X = 2, then p = 2 and Y can only be 3.  (this leads to solution (2,3,2)
For instance, if X = 3, then p = 2 and no Y is possible.
For instance, if X = 4, then p = 3 and Y can only be 5.
...
For instance, if X = 31, then p = 17 and Y can only be 32 or 33.
For instance, if X = 32, then p = 17 and Y can only be 33.
For instance, if X = 33, then p = 17 and no Y is possible.
In fact, this is the case whenever X is 1 more than twice a prime.
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If X = 10,000, then p = 5003 and Y can only be between 10001 and 10005

That's as far as I've gotten, but I suspect a full proof is available by heading in this direction.

 Posted by Steve Herman on 2014-03-23 16:15:55

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