Consider the two equalities:
1!+5! = 11^2
4!+5! = 12^2
Is there a pair of successive integers (below 1000) such that their squares can be written as a sum of 2 factorials?

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

Posted by Ady TZIDON on 2015-04-11 00:46:04