Determine all possible pair(s) (x, y) of nonnegative integers such that (x!*y! - x! - y!) is a perfect square.

{x,y}>4

1. Assume that x,y, are both greater than 4. 
Then x!=10a, y!=10b, for some a,b. 
And z^2=(100ab-10a-10b);z^2 = 10(10ab-a-b) 
But k00 is the only possible square that is a multiple of 10,  
(see list of squares, mod 100: 
00, 01, 04, 09, 16, 21, 24, 25, 29, 36, 41, 44, 49, 56, 61, 64, 69, 76, 81, 84, 89, 96)
So (10k)^2=10(10ab-a-b);10ab = a+b+10k^2 
2. Now we have both 
10a(b-1)-a=10k^2, and 10b(a-1)-b=10k^2 
10a(b-1)-a=10b(a-1)-b; a(10b-11) = b(10a-11)  
And substituting: (x!/10)(y!-11)=(y!/10)(x!-11) 
Which is true iff y=x; but if y=x, then a=b. 
3. Then 10a^2 = 2a+10k^2; 5a^2-a=5k^2 
and a is a fraction, 1/5. 
so either x or y is less than 5, or both are. 

x or y less than 5.

4. Assume y=1 
Then x!-x!-1 = z^2; {x,z}={i,-i}; but that is not a permitted solution. 
5. Assume y=2 
Then (x! -2)=z^2;{x,z}={3,2} 
with all solutions x>=5 ending in 8, which is not possible (see list of squares). 
6. Assume y=3 
Then 5x!-6=z^2;{x,z}={2,2} 
with all solutions x>=5 ending in 94, which is not possible (see list of squares).  
7. Assume y=4 
Then 23 x!-24 = z^2; 23(x!-1) = z^2+1 
Let (x!-1)=k; 23k = z^2+1, but this has no integer solutions. 

So the possible pairs are {2,2}{2,3}{3,2}

Edited on March 29, 2011, 4:19 am

Posted by broll on 2011-03-29 03:06:23