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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    
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solution | Comment 1 of 4
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
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