# Sums Of Successive Squares

Everyone knows of the 3-4-5 triangle. This is the simplest of right triangles with sides measured by integers. It is aften referred to as the Egyptian triangle due to the allegations that the builders of the pyramids might have used it to determine the right angle, alghough of course there are simpler methods to achieve the same goal.

Georges Dostor, at the time professor of theoretical mechanics at the then Catholic University of Paris, published a formula of which 3^2 + 4^2 = 5^2 is a particular case. (3, 4, 5) is the simplest Pythagorean triple. From a different perspective, Dostor's formula equates the sums of a growing number of successive squares: for n \gt 0,

\sum_{i=0}^{n}(2n(n+1)-i)^{2} = \sum_{i=1}^{n}(2n(n+1)+i)^{2}.

The Egyptian triple 3^2 + 4^2 = 5^2 is obtained for n=1. For n=2 the formula gives

10^{2} + 11^{2} + 12^{2} = 13^{2} + 14^{2},

as can easily be verified by hand or with a calculator.

To prove the formula, we just "open" the parentheses and use the well known identity (a\pm{b})^{2} = a^{2}\pm{2ab}+b^{2} .  Observe that most of the terms just cancel out. The number of the terms [2n(n+1)]^{2}   on the left is one more than their number on the right. Additionally, the squares i^{2} on the left are matched one-to-one by the counterparts on the right. Here is what remains

[2n(n+1)]^{2} - 4n(n+1)\sum_{i=1}{i} = 4n(n+1)\sum_{i=1}{i}

which, after dividing by 4n(n+1) and regrouping, reduces to

n(n+1) = 2\sum_{i=1}^{n}{i}

which is a famous formula for triangular numbers that was known yet to the young Gauss.

References

G. J. Dostor, Questions sur les nombres, Archiv der Mathematik und Physik, 64 (1879), 350-352.