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 answer is yes.
The pair given for consideration {11,12} is demonstrated by the assigned equalities to fulfill the requirement.

Besides the given pair {11, 12}, there is also the pair {-12,-11}.

(-12)^2 = 144 = 4!+5!
(-11)^2 = 121 = 1!+5! = 0!+5!

Posted by Dej Mar on 2015-04-07 14:24:56