Post pythagoreas proofs/integer triples you can find!!!!

heres some pythagoream triples:

3,4,5
5 12 13
7 24 25
9 40 41
11 60 61
13 84 85

see if you spot the pattern... (its easy i spotted it in third grade or fourth)

You can assign an ordered pair to each of the triples. Call the pair (x,y). The first number in the triple is x²-y². The second is 2xy. The third is x²+y².

This looks like special binomial products, like (a-b)^2, (a+b)^2, etc.

Also, at least one number in each triple is divisible by 4 and at least one is divisible by 5. I've never been able to prove it, although i know there is one.

So you could do the same thing with 4?

Oh yeah, thanks.

Anyone find proofs of this yet?

Here's a better proof of the fact that at least one element has to be a multiple of 4.

Take the equation x^2 + y^2 = z^2 in modulo 8.

A perfect square is either 1 ( if it is odd) (4 if it is even but not x is not a multiple of 4) or 0(if it is a multiple of 4) modulo 8.

If none of them are 0 modulo 8, Then x^2 + y^2 can either be 1+1 = 2, which is not a value a perfect square can have, 4+1=5(same thing) or 4+4 = 0, which implies z is a multiple of 4. Therfore, one of them has to be a multiple of 4.

i meant proofs ot eht ehorem, not the proofs of my statement earlier :DDD

The thereom that all pythagorean triples can be expressed as (c(x^2-y^2), 2xcy, and c(x^2+y^2))?

Here is a proof.

First, divide gcd(x,y,z) out. This will be c.

After that, one of (x,y) will be odd, and z will be odd. Why? If any two of them are even, all of them are even, and that can't happen 'cause we divide the gcd out. If z is even, then we consider the equation modulo 4, and since squares can only be 0 or 1 modulo 4, then x and y are both eve, contradiction.

Let's say y is even.

Let m be (x+z)/2, and let n be (z-x)/2. Then x= m-n and z = m+n. We get y the square of y is 4mn. If m and n are not relatively prime, then their sum and difference are also not relatively prime. Furthermore, y cannot be relatively prime because y^2 = z^2-x^2. So m and n are relatively prime. Since 4mn is a square and 4 is a square, mn must be a square. Because m and n are relatively prime, m must be a square and n must be a square. Now we get the thereom. QED

15 112 113
16 30 34
32 60 68
64 120 136
17 144 145
19 180 181
21 220 221