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

Home > Just Math
Squares sum up to a prime (Posted on 2017-01-13) Difficulty: 3 of 5
List all integers n such that the sum 2^2 + 3^2 + 4^2 + ... + n^2 equals a prime number. p&p solutions only.

See The Solution Submitted by Ady TZIDON    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
solution | Comment 1 of 6
The sum S = n*(n+1)*(2n+1)/6 - 1 = (n-1)*(2n^2 +5n + 6)/6.

The second factor is always > 1 so for prime S the first factor must reduce to 1.

The possibilities are:
(n-1) = 1, n = 2, S = 4
(n-1)/2 = 1, n = 3, S = 13
(n-1)/3 = 1, n = 4, S = 29
(n-1)/6 = 1, n = 7, S = 139

So S is prime only when n = 3,4, or 7.


  Posted by xdog on 2017-01-13 11:37:51
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (1)
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 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information