# Interesting Math

## Sums of Squares of Consecutive Integers

A couple issues ago, I posed the following question:

We know 32 + 42 = 52; notice 3, 4, 5 is consecutive and there is 1 more term on the left. Can we extend this idea? In particular, does there exist n + 1 consecutive positive integers whose squares sum to the sum of squares of the next n consecutive integers?

This will be a more technical article than usual, but it is such an interesting problem I decided to go over it. If you tried this out, you might have discovered that it certainly holds for n = 2 as well, since 102 + 112 + 122 = 132 + 142 . To solve this problem in general, we want to solve for a such that

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where I have rewritten the sum on the right to make it the difference between "full" sums starting from 0. Rearranging and expanding the squares, this is simply

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You may recall (or rederive) the formulae for sums of consecutive integers or consecutive squares:

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After substituting and simplifying, we end up with

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which is a qudratic in a. Solving, we find two solutions: a = -n and a = n(2n + 1). The first solution makes sense, since we will have -n, -n+1, ... 0 as the first n + 1 consecutive integers, which trivially has sum of squares equal to the sum of squares of the next n consecutive integers. Since I wanted positive integers, the second solution will answer the problem. This is quite an interesting result: for any n, there exist n + 1 consecutive positive integers whose squares sum to the sum of squares of the next n consecutive integers!

Can we extend this result to higher powers? For example, 33 + 43 + 53 = 63. Then in general, does there exist n + 2 consecutive positive integers whose cubes sum to the sum of cubes of the next n consecutive integers? Unfortunately, this is not true. The result fails for the next value of n, n = 2. Indeed, suppose there is an integer a with

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Then this equation holds in every modulus, in particular, in mod 4. However, the first two terms cancel in mod 4, leaving the sum of two consecutive cubes is 0 in mod 4, which never happens (the sum of consecutive powers in mod 4 is either 1 or -1, for any positive power). Oh well, we tried. Maybe there's another generalization to be found — but that's an article for another time.

Vince's problem of the issue

Let S be the set of integers n such that some factorization of n into factors a1,...,an satisfy a1 + ... + an = 2009. What is the maximum element of S?

Vince Chan
v2chan@math.uwaterloo.ca