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=(100ab10a10b);z^2 = 10(10abab)
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(10abab);10ab = a+b+10k^2
2. Now we have both
10a(b1)a=10k^2, and 10b(a1)b=10k^2
10a(b1)a=10b(a1)b; a(10b11) = b(10a11)
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^2a=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 20110329 03:06:23 