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www.mikescher.com/www/statics/euler/Euler_Problem-075_explanation.md
2018-02-03 16:50:58 +01:00

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Luckily there is a simple formula to generate [Pythagorean triples](https://en.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple).
~~~
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).