*[This is a back-issue of one of this site’s newsletters]*

Last week, I showed you how a puzzle about rectangles gives you a simple formula for pythagorean triplets. This week, I want to show you a bit more about that formula.

But first, have you seen this? It’s proof that even people in a marketing career need to know a bit of mathematics:

As someone said: “Once you’ve seen this as a Venn Diagram, it’s hard to un-see”. Poor Reuters!

Ok, back to pythagorean triangles. The formula went like this:

- Start with a number, T
- Work out 1/T
- Then, work out T + 1/T and T – 1/T.
- Divide these both by 2, and put them over a common denominator.
- Then, if (T+1/T)/2 = R/Q and (T-1/T)/2 = P/Q, then P
^{2}+ Q^{2}= R^{2}.

If you make T equal to M/N, then this process gives P=M^{2}-N^{2}, Q=2MN and R=M^{2}+N^{2}. That’s a much more explicit formula, and is known to us from Euclid’s “Elements”, a famous collection of ancient Greek mathematics.

There’s a much simpler way to come up with this formula, which uses complex numbers. Yes, complex numbers make some things simple!

- Start with a complex number M + iN.
- Square it: (M+iN)
^{2}= (M^{2}-N^{2}) + (2MN)i. - Take the absolute value of both sides: |M+iN|
^{2}= |(M^{2}-N^{2}) + (2MN)i|. - Expand out the squared absolute value: M
^{2}+N^{2}= |(M^{2}-N^{2}) + (2MN)i|. - Let P=M
^{2}-N^{2}, Q=2MN and R=M^{2}+N^{2}: R = |P+iQ| - Square both sides of this: R
^{2}= |P+iQ|^{2} - Expand out the last absolute value: R
^{2}= P^{2}+Q^{2}.

So there you have it – Pythagorean triples from complex numbers. Who would have thought!