The trivial answer: yes, there is at least one couple : (-11, -12); it creates the same squares as (11, 12).

On a broader note let us address the absolute values and look for additional couples.

Since 10! is over 1,000,000 and 9!+8! below 1,000,000 we need not to check candidates over 9! .

The difference between adjacent squares is always odd and all the factorials, except 1!, are even numbers, therefore one of the couples must include the number 1! (or 0! which is also 1).

**We are looking for n such that n!+1 is a square**: it works for n in (4,5,7 )==>(25,121,5041).

Checking for the neighbors of 71(the sqrt of 5041), i.e. 70 or 72 creates no solution.

Both 5 and 11 lead to the single (in absolute values) solution i.e.

**1!+5! = 11^2 **

4!+5! = 12^2

BTW, if a 3-way representation were allowed:

**1!+7! = 71^2 **

4!+5!+7! = 72^2

*Edited on ***April 11, 2015, 7:19 am**