p, q and r are three positive integers satisfying 1/p + 1/q = 1/r, such that there is no integer ≥ 2 dividing p, q and r simultaneously. However, p, q and r are not necessarily pairwise coprime .

Prove that (p+q) is a perfect square.

We prove that numbers p,q and r are like that :

p=ab;q=bc;r=ca;

So let consider p=ab and q=bc with a,c coprime

From 1/p+1/q = 1/r we find that

r=pq/(p+q) = ab²c/b(a+c) = abc/(a+c).

But ac and a+c are coprime, so b=a+c and r=ac.

If b=á(a+c) and r=áac then á divide b and á divide r and

will exist a number who will divide p, q and r simultanious!

So b=a+c and p+q = ab+bc = b(a+c) = b² a perfect square!

 

 

Posted by Chesca Ciprian on 2007-09-23 14:35:07