# Pythagorean triples

## 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.

Since any polynomial can be written such as $$p(x) = a_{n}x^{n} + ... + a_{n+1}$$, and since the equality of polynomials theorem holds true, any result obtained with polynomials is just rewritting very simple identities. If n+((n-1)/2)^2=((n+1)/2)^2 holds true, it's because the left and right side simplify to the exact same expression. Sorry to be the bearer of bad news, but you didn't find anything new.

Yes, this is just a trivial consequence of $(a+b)(a-b) = a^2 - b^2$.

There's a semi-interesting related observation that you can make (at least it was semi-interesting to me when I considered it briefly in my first year of undergrad!): What integers $n$ can be expressed as a difference of squares in a nontrivial way - ie. not just using $(n+1)/2$ and $(n-1)/2$?

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rbj
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.
did you try it out? like n=9 or n=25 or n=49?

n=9 -> 3, 4, 5 that one i knew of

n=25 -> 5,12,13 that one i knew also

the next one i might not be familiar with:

n=49 -> 7,24,25 i guess that works.

n=81 -> 9,40,41 i guess this works, too.

n=121 -> 11,60,61

you might not be the first to discover it, but i hadn't known this before (and i'm in my 6th decade, but i never took a course in Number Theory, which is where i'll bet this would be). so i'm more favorably impressed than the others.

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The reason that I say it is trivial is that a = (n+1)/2 and b = (n-1)/2 is precisely the solution to the linear system a-b=1 and a+b=n, for any integer n.

So you're guaranteed to have a^2 - b^2 = (a-b)(a+b) = n. If n is a square that obviously gives you a triple.

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Similar

Hey, what you found is really similar to what I found.
This is my working:
If n is odd: then n, n^2/2 - 1/2 , n^2/2 + 1/2 is a pythagorean triple.
But if n is even: then n, n^2/4 - 1 , n^2/4 + 1is a pythagorean triple.

So I took it to the next level to be able to say this:
n, n^2/a - a/4 , n^2/a + a/4 is a pythagorean triple for ANY value of a, but not all values of a will give all whole numbers. This only works when 2n > a > 0. I also found that if n is even, than if a is divisible by 4 there is a high chance that it will all be integers. Or if n is odd, then a should be even, but NOT divisible by 4.