Thread: Triples -- equation

Seems like a problem solved, but I need help.

The point was to prove that - with relatively prime - had an infinite number of solution in positive integers . I plug to arrive at . Bezout's lemma tells me this has an infinity of solutions and hence an infinity of solutions . But if are of opposite signs, then when rises, decreases, and are not always positive are not always integers.
What am I missing?

You are extremely close to the solution. The last brick in the proof comes from the more explicit version of Bezout's lemma. Given for , not only do there exist an infinity of integer solutions , but they can also be enumerated by

In your equation, find a solution to and call it . You will now notice that for any integer , another solution exists . So for a high enough , and will both be positive, and by increasing arbitrarily, all higher solutions will increase positively without limit.

Originally Posted by Media_Man

... another solution exists . So for a high enough , and will both be positive ...

But here, again, when increases, decreases, provided m and n are opposite signs... right?
Take m=-5,n=8, relatively prime hands down. I naively created a table in Excel to help me find the solutions, but none of them is a pair of positive integers, as they do disperse in different directions...

Ah, I see. In that case, you did not need my help at all. You have already reached the conclusion, that the theorem as you have interpreted it is false. The way it is written, however, is perfectly fine. In my experience, "two relatively prime m,n" implies they are natural numbers, not integers.

Originally Posted by Media_Man

Ah, I see. In that case, you did not need my help at all. You have already reached the conclusion, that the theorem as you have interpreted it is false. The way it is written, however, is perfectly fine. In my experience, "two relatively prime m,n" implies they are natural numbers, not integers.

I can't say it's false, because I have assumed that a,b,c are powers of 2, wich does not need to be the case. And I'm sort of reluctant to overthrow something that must have been checked by people far more experienced than me... I decided relative primes can be negative, as you can write a congruence equation with them etc etc. Anyway, thanks for your help.