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 Some sums (Posted on 2015-04-07)
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?

 Submitted by Ady TZIDON No Rating Solution: (Hide) 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 overv 9! . The difference between adjacent squares is always odd and all the factorials, except 1!, are even – therefore one of the couples must include the number 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 neighbors of 71, i.e. 70 or 72 creates no solution. Both 5 and 11 lead to the single (in absolute values) i.e. 1!+5! = 11^2 4!+5! = 12^2 BTW, if a 3way representation were allowed: 1!+7! = 71^2 4!+5!+7! = 72^2

 Subject Author Date anal. solution..................... spoiler Ady TZIDON 2015-04-11 00:46:04 computer solution Charlie 2015-04-07 14:34:22 answer Dej Mar 2015-04-07 14:24:56

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