Legal Commentary

Math Puzzle

By Eugene Volokh

Here?s a puzzle I came up with this morning (though I?m sure others had thought of it before).

We know that the sum of integers from 1 to n ? let?s call it S₁(n) ? is n × (n+1) / 2.

We know that the sum of the squares of integers from 1 to n ? let?s call it S₂(n) ? is n × (n+1) × (2n+1) / 6.

We know that the sum of the cubes of integers from 1 to n ? let?s call it S₃(n) ? is the square of S₁(n).

Of course the sum of the cubes (S₃) is always divisible by the sum of the integers themselves (S₁).

The sum of the squares (S₂) is sometimes divisible by the sum of the integers (S₁). For instance, consider n=4: 1+4+9+16 = 30, and 1+2+3+4 = 10.

When is the sum of the cubes (S₃) ever divisible by the sum of the squares (S₂), for n > 1?

Full post as published by The Volokh Conspiracy on January 16, 2011 (boomark / email).