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## Main Question or Discussion Point

I developed a formula for finding pythagorean triples recently, I thought I'd share it just for general novelty:

n+((n-1)/2)^2=((n+1)/2)^2

Where n= odd perfect square

a^2+b^2=c^2

Pythagorean triple set is (a,b,c)

So pythagorean triple set for my formula is:

(sqrt(n),((n-1)/2),((n+1)/2))

If it's restating something old or obvious, let me know.

n+((n-1)/2)^2=((n+1)/2)^2

Where n= odd perfect square

a^2+b^2=c^2

Pythagorean triple set is (a,b,c)

So pythagorean triple set for my formula is:

(sqrt(n),((n-1)/2),((n+1)/2))

If it's restating something old or obvious, let me know.