 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Factorial Difference = Perfect Square (Posted on 2011-03-28) Determine all possible pair(s) (x, y) of nonnegative integers such that (x!*y! - x! - y!) is a perfect square.

 No Solution Yet Submitted by K Sengupta Rating: 2.5000 (2 votes) Comments: ( Back to comment list | You must be logged in to post comments.) possible solution Comment 3 of 3 | {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 Please log in:

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