Can you find new pythagorean triangles?

If you are not sure what a pythagorean triangle is, read this first.

The triple 2m+1, 2m(m+1), 2m(m+1)+1 (for all m>0) generates all triples where the two larger numbers differ by 1. Because two of the numbers differ by 1, we know that there can be no divisors of this triple, and so they must all be primitive. How did we find this?

Another triple is 4m, 4m²-1, 4m²+1. m=1 overlaps with the previous series, but all higher m are unique. These all have two of the numbers differing by 2. How did we find this?

Can you generate a similar triple of numbers where one pair of numbers differs by 3?

Let these two numbers be n+3 and n. The difference between their squares is
(n+3)²-n² = 6n+9

For a pythagorean triple, this must also be a square number. Since it is clearly odd, let us set it to (2m+1)².

Then n=((2m+1)²-9)/6 = (2m(m+1)-4)/3, and the triple is 2m+1, (2m(m+1)-4)/3, (2m(m+1)+5)/3

m=1 gives 3,0,3 which doesn't help.

m=2 and 3 give answers in thirds. However, we can still consider 3 times these which will be integers. This suggests that we need to consider 3 cases depending on the remainder when dividing by 3.

Setting m=3k+1 gives 3 times a sequence we have seen before so we can ignore it.

Setting m=3k+2 and multiplying by 3 gives

18k+15, 18k²+30k+8, 18k²+30k+17

These must also be primitive, as the second two numbers differ by 9. Therefore any common factor must be 3 or 9. However neither of these numbers are divisible by 3, they both have remainder 2 when divided by 3.

Setting m=3k and multiplying by 3 gives:

18k+3, 6k(3k+1)-4, 6k(3k+1)+5

These series that we have produced produce primitive triples, but the expressions are getting increasingly cumbersome. Is there a better way? Is there a simpler way? Can we carry on with this program forever?

Yes, we could carry on, but there is a better way. Having started this process with Pythagoras's theorem, you would be surprised if the Greeks didn't take it further. Euclid produced this formula in two integers, m and n.

m²-n², 2mn, m²+n²

It only takes a moment to verify that this satisfies pythagoras's theorem for all values of m and n.

It does produce triples which are not primitive, but it also produces all of the primitive triples.