The Emergency Maths Kit Strikes Again!

Those who know me enjoy laughing at the fact that I have an emergency maths kit in my car (no really!). A simple pad, pencil, sharpener, ruler blah for those moments when I’m sat in my car waiting for my nth child to finish their sporty mcsport face. Such moments, whilst undeniably nerdy, feel pretty important. Like a musician, a mathematician needs to keep practicing, or you get worse at it (at least, I do). So sometimes I prove the area of a trapezium in different ways for fun, or others, I mess around with things floating in my brain like this fun factorial thing I did a while back.

This time, I was fascinated by something I read about consecutive square numbers. It goes  like this:

32 + 42 = 52

102 + 112 + 122 = 132 + 142

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

This is all I had, but it got me thinking. Are there more? Is there a pattern at work? Clearly the next one would involve 9 consecutive numbers, with the first five squares summing to equal the last four squares. Investigate it yourself if you want, or lazily read on!

I set about brute-forcing out the next starting number, by just using a, a+1, a+2 etc for each consecutive number, then solving:

Part 1

The – 4 example is fairly trivial, but the next magic number presented itself : 36.

So I tried it just in case I was wrong somehow:

part 2

Turns out that works, and they both equal 7230. Nice!

I wanted to generalise this pattern, so I started looking at them all together:

part 3

So now I had an nth term to find the first number for each sequence of consecutive squares for this curious pattern. Pretty cool huh? I felt pretty satisfied with it, until I started explaining it in my head, and found the explanation got really messy when trying to decode what you’re supposed to do once you have the starting number.

Back to the notebook!

Each side of the equation is a sum, so I could use Sigma, with limits – as the sum utilises consecutive numbers. The first term is already defined, and then we simply add 1, then 2, then 3, then 4 etc to it – which I declared as a second variable, x. This variable would be zero for the first term, then 1, 2 etc.

The left hand side of the equation had (n+1) terms, and the right hand side of the equation always has n terms. Hence in total there are (2n+1) terms.

Therefore x, starting at zero for the first term, needs to increment by 1 up to a limit of ‘n’ on the left hand side of the equation (0 to n is the same number of increments as n+1), and start where we left off on the right hand side, ie, n+1, up to a limit of 2n.

All together then:

z5xOJ6mJ

Or if you want a prettier version:

general

Could make for an interesting investigation for students!