how many pathagorean triples have legs between 1 and 500?

Answer 1

let d be the difference between hypotenuse and longer leg.

d=1, (3,4,5) - (31,480,481) => 15 (every other 2)
d=2, (6,8,10) - (44,483,485) => 20 (every other 2)
d=3, (9,12,15) - (51,432,435) => 8 (every other 6)
d=4, (12,16,20) - (60,448,452) => 13 (every other 4)
d=5, (15,20,25) - (65,420,425) => 6 (every other 10)
d=6, (18,24,30) - (72,429,435) => 10 (every other 6)
d=7, (21,28,35) - (77,420,427) => 5 (every other 14)
d=8, (20,21,29) - (88,480,488) => 18 (every other 4)
d=9, (27,36,45) - (93,476,485) => 12 (every other 6)
d=10, (30,40,50) - (100,495,505) => 8 (every other 10)
d=11, 33 - 99 => 4 (every other 22)
d=12, 36 - 108 => 7 (every other 12)
d=13, 39 - 91 => 3 (every other 26)
d=14, 42 - 112 => 6 (every other 14)
d=15, 45 - 105 => 5 (every other 15)
d=16, 40 - 120 => 11 (every other 8)
d=17, 51 - 119 => 3 (every other 34)
d=18, 48 - 132 => 15 (every other 6)
...
d=125, 375 => 1

so you have to do this up to d=125, but i only have the energy until d=18. my incomplete sum is 169. it could be nearly 1000 if i'm not mistaken.

Answer 2

Every integer from 3 on up is part of at least one Pythagorean triple.

The reason for that is that for any odd integer n:

n, (n^2-1)/2, (n^2+1)/2 is a Pythagorean triple.

And of course there's then a triple starting with 2n as well.

That doesn't exactly give you a count of the triples, but it should at least give you a sense of the magnitude of the answer.

I once worked out, just for fun, an exact formula for generating all Pythagorean triples. I think the same formula is widely available on the web, if you do an obvious web search. Such a formula would indeed guide you to an exact count, if you really need one.