if $(x+y)(x-a)(y-a)=n$ has a solution at $x=0$, then $x \geq 0$

Question

Is this true? That if:

$(x+y)(x-a)(y-a)=n$ $\hspace{1cm}$(where $a>0$ and $n>0, x,y\in \mathbb{Z}$)

has a solution when $x=0$ then other solutions occur only when $x\geq0$. If it is true then why and how to prove it?

For example if $a=7$ and $n=42$ the only solutions are $(0,1), (0,6), (1,6)$ and their reflections, so no negative solutions. [and when I say reflection I mean $(1,6)\equiv(6,1)$]

And if $a=14$ and $n=72$ the solutions are $(11,13), (-10, 11), (-10, 13)$ and their reflections but there is no solution when $x=0$.

To state the question another way: Is it true that if there is not a solution when $x=0$, and a solution exists, then there must be a solution where $x<0$?

(I am not sure if I need to define $a,n\in \mathbb{Z}$ also)

Answer 1

In a similar vein to the excellent answer by Raffaele:

Given positive integers $a$ and $n$, consider the equation $$(x+y)(x-a)(y-a)=n.$$ If it has an integral solution $(x,y)\in\Bbb{Z}^2$ with $x=0$ then $$n=(0+y)(0-a)(y-a)=-ay(y-a),$$ where $y-a\neq0$ because $n>0$. It follows that $$-ay(y-a)=n=(x+y)(x-a)(x-a),$$ where we can divide by $y-a$ and rearrange a bit to find that $$x(x+y-a)=0.$$ So for any solution $(x,y)$ with $x\neq0$ we must have $x+y=a$. Then $$n=(x+y)(x-a)(y-a)=axy,$$ which shows that $xy=\tfrac{n}{a}$ and $x+y=a$. In particular the sum $x+y$ and the product $xy$ are both positive. This implies that both $x$ and $y$ are positive.

Answer 2

$$(x-a) (y-a) (x+y)=n$$ has solution $x=0$ means that $$-a y (y-a)=n \quad (1)$$ substitute in the first equation $$(x-a) (y-a) (x+y)=-a y (y-a)$$ expand $$a^2 x-a x^2-2 a x y+x^2 y+x y^2=0$$ and factor $$x (y-a) (x+y-a)=0$$ the other solution is $$x = a - y$$ it is given that $a>0$ so we can conclude that $x\ge 0$ only if $y\le a$.

Hope this is useful

edit

from equation $(1)$ we get that $$y=\frac{1}{2} \left(a\pm \sqrt{a^2-\frac{4 n}{a}}\right)$$ in order to be real we must have first of all $$n\geq \frac{a^3}{4}$$ If we want integer $y$, further condition must be verified.