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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.

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    $\begingroup$ Check whether solutions would be odd or even $\endgroup$ – Empy2 Dec 18 '19 at 12:27
  • $\begingroup$ How could I go about that? $\endgroup$ – Rodrigo Dec 18 '19 at 12:28
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    $\begingroup$ There are four options, one of which is 'x is even, y is odd'. In that case,would the left-hand side be odd or even ? $\endgroup$ – Empy2 Dec 18 '19 at 12:30
  • $\begingroup$ It would always be odd (or reduced to be odd), so it couldn't equal zero? $\endgroup$ – Rodrigo Dec 18 '19 at 12:37
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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.

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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.

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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.

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  • $\begingroup$ So, since $y^2-xy-x^2$ can always be reduced to be odd, it cannot equal 0 (an even number)? $\endgroup$ – Rodrigo Dec 18 '19 at 12:36
  • $\begingroup$ it shows that the only solution is $(0,0)$ since if $(x,y)$ not zero, after a finite number of steps, $x$ or $y$ will be odd. $\endgroup$ – Tsemo Aristide Dec 18 '19 at 12:39
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    $\begingroup$ @Rodrigo Let me explain Tsemo Aristide's solution. It is easy to see that if $x$ and $y$ have differrent parity or both $x$ and $y$ are odd, then the equation $y^2 - xy - x^2 = 0$ does not hold. So, it must be that both $x$ and $y$ are even. Now, if $x$ and $y$ are solutions, so is $(\frac{x}{2},\frac{y}{2})$ as shown in solution above. So, we keep dividing $2$ to the solution. Eventually at least one of them must be odd. We know that if one of solutions is odd, then there is a contradiction. Thus, we conclude that there is no positive integer solution to $y^2 - xy - x^2 = 0.$ $\endgroup$ – Idonknow Dec 18 '19 at 12:46
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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:

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

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  • $\begingroup$ Where does the \Delta come from? Why does it hold up? Are you just taking $x$ or $y$ (ala partial derivatives) to be constant and solving a quadratic equation on just one variable? $\endgroup$ – Rodrigo Dec 18 '19 at 12:46
  • $\begingroup$ Yes, they are looking at the discriminant of the equation, considered as a quadratic in $y$. Where they say integer, they mean "nonzero integer". $\endgroup$ – Andrés E. Caicedo Dec 18 '19 at 12:48
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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.

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