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# Formaula for pythagorian triples watch

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1. ive been thinking bout this recently and i belive there are a infinate number, and also ive been seeing similarities in them as they get bigger,

any1 else have ny thoughts on this
2. It can be shown that, the three sides of a right angled triangle can be written in the form

p^2-q^2
p^2+q^2
2pq

And yes, there are similarities. For example, you have a 3,4,5 triangle, along with a 6,8,10 one

so if a^2+b^2=c^2
and you double a and b, and let the sum of their squares be x
you get 4a^2+4b^2=x^2
a^2+b^2=c^2, x^2/4=c^2
so x=2c

There is also quite obvious stuff, such that, they are either all even or only one is even. But that is obvious from the fact that an odd+odd=even, and even+odd=odd and even+even=even
3. The product of a pythagorean triple is a mutltiple of 60.
4. if the proof difficult? ^
I tried showing that the square of the product [(a^4)(b^2) + (a^2)(b^4)] is a multiple of 3600. But I'm not sure cancelling c is a good idea because the set of numbers have to follow the rule and can't just be any 3 integers.
5. (Original post by lgs98jonee)
It can be shown that, the three sides of a right angled triangle can be written in the form

p^2-q^2
p^2+q^2
2pq

There is also quite obvious stuff, such that, they are either all even or only one is even. But that is obvious from the fact that an odd+odd=even, and even+odd=odd and even+even=even
Yes, one must be even and one must be odd, but thats not where the p^2 - q^2, p^2 + q^2, and 2pq come from. In fact, this parameterization requires quite a bit of work. Its interesting, however, because its literally half the proof that Fermat used in his infinite descent argument for n=4.
6. (Original post by mik1a)
if the proof difficult? ^
I tried showing that the square of the product [(a^4)(b^2) + (a^2)(b^4)] is a multiple of 3600. But I'm not sure cancelling c is a good idea because the set of numbers have to follow the rule and can't just be any 3 integers.
If you know a bit of modular arithmetic then it's not too bad.

Square numbers can be 0 or 1 mod 3, can be 0 or 1 mod 4, and can be 0, 1 or 4 mod 5.

So looking at a^2 + b^2 = c^2 we see that:

mod 3 this eqn has to be 0 + 1 = 1 or 0 + 0 = 0;
mod 4 this eqn has to be 0 + 1 = 1 or 0 + 0 = 0;
mod 5 this eqn has to be 0 + 0 = 0 or 0 + 1 = 1 or 1 + 4 = 0 or 0+4=4.

Any which way there has to be factors of 3,4 and 5 in abc and so the product is divisible by 60.

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