Showing a diophantine equation has no solutions, $y^2-xy-x^2=0$.

Question

I'd like to show the following equation doesn't have any positive integer solutions.

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

How can I show that said equation doesn't have any solutions in the set of positive Integers?

I've tried factoring it out, and manipulating the equation to no avail.

Answer 1

Suppose $y^2-xy-x^2=0$ for some $x,y \in \mathbb Z^+$.

then, $(\frac{y}{x})^2-\frac{y}{x}-1=0$

thus, $\frac{y}{x}=\frac{1 \pm \sqrt{5}}{2}$ which contradicts the supposition.

Answer 2

If $y$ odd or $x$ even, $y^2-xy-x^2$ is odd.

$x,y$ even $x=2a, y=2b, (2b)^2-4ab-(2a)^2=0$ implies that $b^2-ab-a^2=0$ you reduce until $a$ or $b$ is odd.

Answer 3

We simply solve the equation for y or x, considering x or y any arbitrary integer:

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

$\Delta=x^2+4x^2=5x^2$

$\sqrt {5}$ is not integer therefore y is not integer if x is integer. The same is true for x when y is any arbitrary integer, because:

$\Delta= y^2+4y^2=5y^2$.

Answer 4

Here is a different approach, based on the idea of completing the square. First, a number is 0 if and only if four times the number is 0, so we want to check that $4y^2-4xy-4x^2=0$ has no positive integer solutions, that is, we want to show that $$ 5y^2=(2x+y)^2 $$ has no positive integer solutions. But this is clear: if $y\ne0$ (which implies that $2x+y\ne0$), then 5 appears in the prime factorization of the number on the left an odd number of times while it appears an even number of times in the prime factorization of the number on the right. So $y=0$, and therefore $2x+y=0$, so $x=0$ as well.

Answer 5

Assume we have a solution $x,y$ with smallest $x$. Inspecting the equation $\bmod 2$, we can see that only $x\equiv y \equiv 0 \mod 2$, so $x=2m$, $y=2n$ for some positive integers $m,n$. Putting this back to the equation we get $4m^2-4mn-4n^2=0$, and so $m^2-mn-n^2=0$ with $m<x$, a contradiction.