Sea shells of Maths

I have loved Mathematics ever since I can remember - its interesting patterns, astonishing results, and elegant proofs. As a kid, I spent a lot of time playing with numbers, finding more and more silly patterns, and looking for the rules that would explain them. I would spend hours at end on my explorations, like a child trying to collect the prettiest sea shells on the beach.

The more I learnt, the more I wanted to explore, but I never really dived into the depths; I just waded as deep as my basic education would let me. And to this day, I'm still the child who gets excited by pretty sea shells on the shore of Mathematics. Let me show you a typical sea shell I found the other day, but first I'll tell you the story of how I found it.

It's a long story, which begins when I was checking some calculations and computed, among other things, the squares of 119 and 169. I noticed that 119² + 120² = 169², and I started thinking about other such examples.

One generalization is of course Pythagorean triplets

a² + b² = c²

and we already know how to generate them using

a = m² - n², b = 2mn, c = m² + n²

For m = 12, n = 5, we get our equation above.

But I was now considering another possible generalization, of this form:

Sum of (m + 1) consecutive squares on the left side equals sum of m consecutive squares on the right. Some examples:

For m = 1:

20² + 21² = 29²

119² + 120² = 169²

For m = 2:

108² + 109² + 110² = 133² + 134²

For m = 3:

21² + 22² + 23² + 24² = 25² + 26² + 27²

A further generalization is to take squares with a constant difference d, instead of consecutive squares. Some examples:

For d = 2, m = 3:

9² + 11² + 13² + 15² = 12² + 14² + 16²

165² + 167² + 169² + 171² = 192² + 194² + 196²

For d = 5, m = 2:

8² + 13² + 18² = 14² + 19²

A little investigation shows that given any m and d, there is a solution for every integer p such that m(m + 1)p(p + d) is a square. For the case where d = 1 i.e. for consecutive squares, this means that given any m, there is a solution for every integer p such that the product of m^th and p^th Triangular numbers is a Square. How pretty is that!

We get the following method to generate the actual solution which, by the way, is how I found the examples above:

Let r = p(m + 1), s = m(p + d), and z = gcd(r, s)
Let a = sqrt(r/z), b = sqrt(s/z)
Let k = zb(a + b) - md, n = za(a + b)

Then the sum of squares of the (m + 1) numbers

k, k + d, k + 2d, ..., k + md

equals the sum of squares of the m numbers

n + d, n + 2d, ..., n + md

So, for example, d = 2, m = 15, p = 3 gives us that the sum of squares of odd numbers from 105 to 135 equals the sum of squares of even numbers from 110 to 138.
Also, d = 1, m = 1, p = 49 gives us the equation we started with, 119² + 120² = 169².

And this finally brings us to the sea shell that I was talking about.
For d = 1, and for any m, putting p = m gives us:

The sum of first (m + 1) squares starting at m(2m + 1) equals the sum of next m squares.

Thus, for m = 1, we get

3² + 4² = 5²

For m = 2,

10² + 11² + 12² = 13² + 14²

For m = 3,

21² + 22² + 23² + 24² = 25² + 26² + 27²

For m = 4,

36² + 37² + 38² + 39² + 40² = 41² + 42² + 43² + 44²

For m = 5,

55² + 56² + 57² + 58² + 59² + 60² = 61² + 62² + 63² + 64² + 65²

And so on, ad infinitum.
That's a nice little sparkling sea shell, isn't it? :)

PostScript

While looking at the equation 119² + 120² = 169², I remembered that 8² + 15² = 17², and I realized that I could rewrite the above equation in a more pleasing form, as

(7²x8²) + (8²x15²) + (15²x7²) = (13²x13²)

Why do I like it? Because it is of the form

(a x b) + (b x c) + (c x a) = (d x d)

where a, b, c, and d are squares of integers! The left hand side of this equation has a sweet symmetry, and I'm a total dope for symmetry. A friend pointed out that we could just take square root on both sides, and get

7² + 8² + 7x8 = 13²

and that using the same idea as above, we can actually write

7² + 8² + 7x8 = 3² + 4² + (3x4)²

which was unexpected to me, and not as pleasing; I felt that it was not as aesthetically neat as the previous one. Beauty lies, my friend reminded me, in the eyes of the beholder.