# Number Theory

• June 8th 2013, 03:58 AM
swarna
Number Theory
There do not exist positive integers x,y,z satisfying 2xz=y^2 & x+z=997
• June 14th 2013, 02:36 PM
Phantasma
Re: Number Theory
Is there a...question?
• June 14th 2013, 07:54 PM
ibdutt
Re: Number Theory
Proceed by using these two equations to form a quadratic equation. Since we have two equations and three variables we cannot have a unique solution. we will get infinitely many solutions.
from first one we have z = (y^2)/(2x) plug in this in equation 2, we get
x + (y^2)/(2x) = 997
Form this quadratic and then try and reason out
• June 15th 2013, 02:49 AM
Idea
Re: Number Theory
Quote:

Originally Posted by swarna
There do not exist positive integers x,y,z satisfying 2xz=y^2 & x+z=997

Suppose x is even
$x=2t$
$4\text{tz}=y^2$
$t=u^2 , z=v^2$ since t and z are relatively prime
$2u^2+v^2=997$ which is impossible (Use congruence mod 8)