Problem 52834. Easy Sequences 32: Almost Pythagorean Triples

Like David Hill, my answer matches the test cases where n<=10, but not for larger ones. I've been driving myself batty the last couple of days trying to figure out where I went wrong. Then I realized that the test cases have it wrong. I can't prove I'm right, but I can prove them wrong:
Consider the defining condition: c² = a² + b² - 1.
Rearrange: c² - a² - b² = -1.
Modulo 2: c² - a² - b² ≡ 1 (mod2).
As squaring and negation preserve parity (value mod 2),
c + a + b ≡ 1 (mod2)
And all of the answers in the test cases are even for n>10.

My "double" solution matches my int64 solution up through 1000 but not 10,000, which is what I expected as flintmax is ~9e15.. My double solution (generates even perimeters for 10k+) is size 59, my int64 is 63.
I didn't run calculateSize() before trying to figure out what was up, but what I learned through that definitely shaved some points. The largest single value test case(~12e5) runs on my laptop in just under 200ms, and should run in O(n) time and constant memory. It will start overflowing when n hits about 45e5. uint64 will take it to about 9e6.

@Ramon Villamangca, there's definitely an issue with the larger test cases in this problem. From the definition and tracking parity, the perimeter of an APT MUST be odd.