Pythagorean triples

  • #1

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.
 

Answers and Replies

  • #2
1,425
1
Since any polynomial can be written such as [tex]p(x) = a_{n}x^{n} + ... + a_{n+1} [/tex], 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.
 
  • #3
998
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Yes, this is just a trivial consequence of [itex](a+b)(a-b) = a^2 - b^2[/itex].

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 [itex]n[/itex] can be expressed as a difference of squares in a nontrivial way - ie. not just using [itex](n+1)/2[/itex] and [itex](n-1)/2[/itex]?
 
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  • #4
rbj
2,226
7
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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  • #5
998
0
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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  • #6
1
0
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.

What do you think about this?

I went and tested this too. It works for ALL pythagorean triples with the legs smaller than 180000, which kind of gives me a good feeling about it. So I proved it works for a pythagorean triple (it fits into the pythagorean formula), but I want to prove it works for ALL pythagorean triples. Do you have any ideas?

P.S. If this turns out big, I would like to hold rights to it :D
 

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