# 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 $x^2 +xy-y^2 = 0$.

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Have you tried polar coordinates? –  Pragabhava Oct 25 '12 at 2:22
I don't have a solution in polar coordinates. –  mathberry Oct 25 '12 at 13:29
The one problem I saw on this thread is that everyone is giving full answers instead of hints... this question is clearly of the homework type, and the OP won't really learn if you feed him the details. –  Chris Gerig Oct 27 '12 at 18:08

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.

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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.

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Quadratic formula on $x$ treating $y$ as a constant should do the trick! You'll end up with a $\sqrt{5}$ factor that can't disappear.

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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

$$x^2+xy-y^2=x^2+x^2-x^2=x^2=0$$

$$x=0$$

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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.