# When does $(xz+1)(yz+1)=az^{3}+1$ have finitely many solutions in positive integers?

Consider the diophantine equation in three variables $$x$$, $$y$$ and $$z$$; ($$xz+1$$)($$yz+1$$) $$=$$ $$6z^{3}+1$$. The only positive integer solutions I have found are {$$x=4,y=10,z=7$$} and {$$x=10,y=4,z=7$$}. From a Maple program, I have iterated over all values of $$z$$ in the range $$50, the only corresponding solutions of $$x$$ and $$y$$ are those with $$x=0$$ and $$y$$ positive and vise versa. I would like to find out if this diophantine equation contains finitely many or infinitely many solutions in positive integers $$x, y$$ and $$z$$. In general; For a given positive integer $$a$$, what conditions are sufficient for the diophantine equation ($$xz+1$$)($$yz+1$$) $$=$$ $$az^{3}+1$$ to have finitely many solutions in positive integers $$x, y$$ and $$z$$. From experimental results, it appears that this equation has finitely many solutions in positive integers if and only if $$a$$ is not a third power of any integer i.e. $$a\neq m^{3}$$ for all integers $$m$$. Any help or references on this question will be appreciated.

• Note that this is the same as solving $az^3-xz-yz-xyz^2=z(ax^2-x-y-xyz)=0$; if $z>0$, then this reduces to solving $az^2 = x+y+xyz$. Jul 14, 2020 at 1:11
• Thank you Rogerl. Let me try to move from this step and see what conditions are sufficient for $az^{2}=x+y+xyz$ to have finitely many positive integer solutions in $x, y$ and $z$.
– ASP
Jul 14, 2020 at 4:38
• I recommend changing the title to "When does $(xz+1)(yz+1)=az^{3}+1$ have finitely many solutions in positive integers?". The current title asks a question which is too general. I think the current question body provides sufficient context for the suggested title. Jul 14, 2020 at 13:17
• Recommendation has been reflected in the title.
– ASP
Jul 14, 2020 at 13:38
• Anybody here? I have verified using Maple that $(xz+1)(yz+1)=6z^{3} +1$ has no other solutions in positive integers for all $z<10^{8}$ except those solutions already listed in the question. Isn't there some way we can really prove that the positive solutions of this diophantine equation are finite? From the Maple program, I have also iterated over all values of $z$ in the range $50<z<10^{8}$, the only corresponding solutions of $x$ and $y$ are those with $x=0$ and $y$ positive and vise versa. Can't we prove that this will be the trend for all $z>50$ ?
– ASP
Jul 15, 2020 at 6:45

Here is a partial answer: If $$a=b^3$$ is a cube, then there is an infinite family of solutions to $$(xz+1)(yz+1)=az^3+1=b^3z^3+1$$ given by $$(x,y,z) = (b, b^2z-b, z),\ b, z\in\mathbb{N}.$$ This arises from the factorization $$b^3z^3+1 = (bz+1)(b^2z^2-bz+1) = (bz+1)((b^2z-b)z+1)$$.

In addition to the above, for any $$a$$ there are solutions $$(a+1, a^2+a-1, a^2+2a)$$ and $$(2a-1, 2a+1, 4a)$$, and there appear to be (empirically) solutions for some values of $$x$$ between $$a+1$$ and $$2a-1$$. For all of these solutions, $$z = x+y$$ and each $$x$$ corresponds to a unique $$y$$. There appear to be no solutions for $$y\ge x>2a-1$$. These, together with a finite set of solutions for $$x, appear to cover all solutions to the equation. I can prove very little of this.

• I see, this establishes the condition for an infinite number of positive integer solutions. Now the harder part, does the converse hold as well i.e. If the diophantine equation has infinitely many positive integral solutions then $a$ is a third power of some integer?
– ASP
Jul 14, 2020 at 4:45
• I would definitely upvote this answer but am not yet able to upvote. I hadn't noticed these two solutions for every integer $a$.
– ASP
Jul 15, 2020 at 8:56
• It appears there are no solutions in positive integers $x$ and $y$ for all $z>a^{2} +2a$
– ASP
Jul 15, 2020 at 11:31
• You mean if $a$ is not a cube; that is correct. Also, if $x>a$, the minimum value of $z$ appears to be $4a$, given by the solution above. Jul 15, 2020 at 12:10
• Yes, I meant when $a$ is not a cube. When $a$ is a cube, you already showed there are infinitely many solutions in positive integers.
– ASP
Jul 15, 2020 at 12:45