All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers
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?

See The Solution Submitted by Ady TZIDON    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution answer | Comment 1 of 3
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
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information