# Desperately seeking Pythagoras

Also known as Project Euler problem 39:

If p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120.

{20,48,52}, {24,45,51}, {30,40,50}

For which value of p ≤ 1000, is the number of solutions maximised?

My first thought here was to generate an iterable collection of right angled triangles and accumulate a count of triangles by perimeter size. Don’t try this – it is far too slow an approach.

So my second thought was to try and iterate over the perimeter sizes and calculate the number of right angle triangles for each. It’s a long time since I had to think about Pythagoras but, for a triangle with sides `a`, `b` and `c` and a perimeter of `p`, I can make the following two statements:

`a + b + c = p`
`a^2 + b^2 = c^2`

Based on the second statement, above, I can also assert:
`c = sqrt(a^2 + b^2)`

Also:
`c = p - a - b`

Which means:
`a^2 + b^2 = (p - a - b)^2`

This can be expanded and simplified to state:
`0 = pp - 2pb - 2ap + 2ab`

or:

`b = p(p - 2a) / 2(p - a)`

Using this formula, we can say that for any values of `p` and `a` where `p > a`, if `b` is an integer then we have a right angle triangle.

One further optimisation: Intuitively, `a` can’t be greater than `p/3`.

Implementing this can then be done with two very simple `for` loops. The outer one iterates over all values of `p` and the inner one iterates over the range `(2, int(p / 3))` and counts the number of integer values of `b`.