Show $x^2 +xy-y^2 = 0$ is only true when $x$ & $y$ are zero. Show that it is impossible to find non-zero integers $x$ and $y$ satisfying $x^2 +xy-y^2 = 0$.
 A: If $x$ and $y$ are nonzero integers, $y=qx$ for some rational number $q$. This gives $$x^2+qx^2-q^2x^2=0\Rightarrow q^2=q+1$$ which is not solved by any rational number q.
A: If $x$ and $y$ are both odd, or if one is even and the other is odd, then the left-hand side is odd and can't equal $0$.  So any solution must have $x$ and $y$ even.  But if there is any nonzero integer solution $(x,y)$, then $(x/k, y/k)$ is also an integer solution, where $k$ is the highest power of $2$ dividing both $x$ and $y$.  This reduced solution would then have either $x/k$ or $y/k$ odd, which cannot happen.  Hence there can be no nonzero integer solution.
A: Let $g=\gcd(x,y)$.  Let $x=rg$ and $y=sg$.  Then
$$g^2r^2+g^2rs-g^2s^2=0$$
$$r^2+rs-s^2=0$$
$$s^2-r^2=rs$$
$$(s+r)(s-r)=rs$$
We've divided out all common factors, so $\gcd(r,s)=1$.  For any prime $p$, if $p|r$, $p$ does not divide $s$ and therefore divides neither $s+r$ nor $s-r$.  The same argument applies if $p|s$.  Therefore, $r$ and $s$ contain no prime factors and $r=s=1.$  Since $r=s$, $x=y$. So
$$x^2+xy-y^2=x^2+x^2-x^2=x^2=0$$
$$x=0$$
A: The quadratic form factors into a product of lines $$0 = x^2 + xy - y^2 = -(y-\tfrac{1-\sqrt{5}}{2}x)(y-\tfrac{1+\sqrt{5}}{2}x),$$ equality holds if either


*

*$y=\tfrac{1-\sqrt{5}}{2}x$

*$y=\tfrac{1+\sqrt{5}}{2}x$


but this can't happen for $x,y$ integers unless they're both zero.
