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Some sums (Posted on 2015-04-07) Difficulty: 3 of 5
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    
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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

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Solutionanal. solution..................... spoilerAdy TZIDON2015-04-11 00:46:04
Solutioncomputer solutionCharlie2015-04-07 14:34:22
SolutionanswerDej Mar2015-04-07 14:24:56
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