# Nontrivial Solutions

• Aug 5th 2009, 08:52 AM
diddledabble
Nontrivial Solutions
Show that $\displaystyle x^2+y^2=4xy$ has no notrivial solutions. I am not sure where to begin. Is it like a Fermat Pell equation?
• Aug 5th 2009, 10:53 AM
Bruno J.
If it has a nontrivial solution, it clearly has one with $\displaystyle x,y$ coprime. But, for example, $\displaystyle y^2=4xy-x^2$ implies that $\displaystyle x$ divides $\displaystyle y^2$... can you finish?
• Aug 5th 2009, 11:23 AM
diddledabble
No, you lost me. Should I rewrite as $\displaystyle x^2+x^2-4xy=4xy$? Then $\displaystyle 2x^2=8xy$
• Aug 5th 2009, 12:16 PM
Gamma
No, those two equations are not the same.

What he is saying get the original equation in the form he has it (subtract $\displaystyle x^2$ from both sides. Then you see the RHS you can factor out an x to see that x must divide $\displaystyle y^2$. But if $\displaystyle x|y^2$, then you see that x and y must share at least 1 prime factor, or else x is $\displaystyle \pm 1$ or y is 0. y=0 yields the trivial solution. $\displaystyle x=\pm 1$ has no integer solutions, use the quadratic formula to see why (or rational root test if you want). So they must share a prime factor making them not relatively prime.

Very nice proof bruno.