www.mikescher.com/www/statics/euler/Euler_Problem-075_explanation.md

Luckily there is a simple formula to generate Pythagorean triples.

a = k * (m*m - n*n);
b = k * (2*m*n);
c = k * (m*m + n*n);

We have a cache of the size 1500001 the we go through all the values for m and n where 2*m*m + 2*m*n <= LIMIT.

When we have found a triple we look in the cache at position a+b+c:

• If the value is not set (0) we write the the triple-hash ((a*7 + b)*5 + c) at the position and increment the result.
• If there is already an hash that equals with our current triple we do nothing
• If there is already an different hash we write -1 in the cache and decrement the result.
• If there is an -1 in the cache we do nothing

So at the end we have the amount of sum's with exactly one solution in our result-value.

We have to test for equal hashes because its possible to find the same tuple multiple times (with different k values).