### Project Euler: Problem 9

After a short break because of Blogger being in read-only mode for quite a while, I'm back again, now with problem No. 9

A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,

a2 + b2 = c2

For example, 32 + 42 = 9 + 16 = 25 = 52.

There exists exactly one Pythagorean triplet for which `a + b + c` = 1000.
Find the product `abc`.

So, here we are, we have the sum and we know that `a < b < c` and on paper the sides would make a right-angled triangle. That gives us `a + b > c` as we can't have a triangle where the sum of the lenghts of two of its sides would be smaller than the length of the third one. For this case it means that `c` can't be bigger than `500`, and as it is as well the biggest of the three numbers it also implies that neither `a` nor `b` can get bigger than `500`. This means that we can set our upper limit to `nr/2`. As `a < b` it makes sense to start searching for `b` at `a + 1`. For each such pair of `a` and `b` we get `c` by substracting them from the target sum and if it works in the Pythagorean equation, we return it.

````fn getTripleForSum nr =`
`(`
`    local half = nr/2`
`    local a, b, c`

`    for a = 1 to half do`
`        for b = (a + 1) to half do`
`        (`
`            c = nr - a - b`
`            if c^2 == (a^2 + b^2) do`
`                return a*b*c`
`        )`
`)`
```

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