Question

How do you prove that if n ≥ 3 is odd, then there exists a pythagorean triple containing n?

Answer 1

Just an idea: You only need to prove it for the prime numbers, because the the factors will already have been proven. Not sure if this helps in the proof.

Answer 2

Ok, I'll try looking at it from that perspective.

Answer 3

(and of course I'm assuming some kind of proof by induction)

Answer 4

Ok :)

Answer 5

perhaps some of these help:
http://en.wikipedia.org/wiki/Pythagorean_triple#Elementary_properties_of_primitive_Pythagorean_triples

Answer 6

"Every integer greater than 2 is part of a primitive or non-primitive Pythagorean triple, for example, the integers 6, 10, 14, and 18 are not part of primitive triples, but are part of the non-primitive triples 6, 8, 10; 14, 48, 50 and 18, 80, 82."

So we can forget about the number being odd, since apparently _all_ numbers greater than 2 are part of a Pythagorean triple.

Answer 7

(perhaps proving the more general theorem is easier)

Answer 8

Okay, I found a proof.

Here are some hints:
There is no induction involved, it is straightforward algebra. Restricting the proof to odd numbers makes the proof easier. And prime numbers also make it easier.

Rather than me just giving you the proof, try and let me know how far you got, and I can give you hints along the way.

Answer 9

Pythagoras: a^2 + b^2 = c^2
Let a be a prime number, then you need to find a way to get any b and c that are also natural numbers.

Answer 10

Let me know how far you get with that.

Answer 11

hello? @Xylienda?

Answer 12

Actually, you don't even need to assume prime numbers; it just helped with the intuition.

Just pick c to be (a^2+1)/2 and b to be (a^2-1)/2 (or perhaps the other way round, I don't remember exactly), then
a^2 = c^2 - b^2
= (c+b)(c-b)

and work from there.

But I guess the OP has disappeared.