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.

Consider 1/6 + 1/3 = 1/2 as an example for n=3.
Then 1/12+ 1/4 = 1/3 is for n=4.

 
Thus q=n, r=(n-1), and p=n(n-1).
 
Usind the LCD as  n(n-1)
 
1/n(n-1) + (n-1)/n(n-1) = n/n(n-1)
 
The next term is for (n+1)
 
1/(n+1)n + n/(n+1)n = (n+1)/(n+1)n
 
Therefore by induction:
 
(p+q) = (n^2 - n) + (n) = (n^2)
 
QED,  CWG

Posted by C W Gardner on 2007-09-27 19:26:47