Oh gosh. That seems really hard. So:

(My super-smart twin brother had to try to figure this one out himself and tell me the answer, this is him talking)

If you take any three consecutive numbers a, b, and c (least to greatest), the difference of [(c^2-b^2)-(b^2-a^2)] will always be 2, and both differences (before subtracting them) will be odd. I don't know how to prove this but I could probably figure it out if I tried but I'm too lazy, and every combination I've tried has worked. I originally thought of this because I knew 5, 12, and 13 was a triple and also 7, 24, and 25 was, so basically it seemed like a pattern. Going back, you can see that every odd number will be represented with this, because all odd numbers are 2 apart. Because all odd numbers when squared make another odd number, we can say that all odd numbers can be done this way as a pythagorean triple.

Knowing that all odd numbers will work, it's time to work on the evens. We know that every multiple of 4 will work, because 3, 4, 5 is a triple and you can just keep multiplying that by whatever factor you need to get a certain multiple of four. So now, the only numbers that aren't accounted for are multiples of 2. The way I thought of it is that every multiple of two that isn't a multiple of four is the product of 2 times any odd number. Because we know that all odd numbers are pythagorean triples, you will always be able to basically scale the odd-numbered triple by a factor of two and get the even number to be in a triple.

Lets take an example: say I want to find a pythagorean triple involving 38. 38 is the product of 2 and 19. Since 19^2 is 361, all I have to do is guess and check until I have narrowed down the right consecutive numbers in the triple. After about a minute, I have found that 181^2-180^2=361, aka 19^2.

So the pythagorean triple involving 19 is 19, 180, and 181. Now we can just double that to get the triple for 38, which would be 38, 360, and 362.

Challenge Accepted. I am the smarter brother, you know.