# Integer points on a surface

I would like to understand $${\bf nonnegative \ integral}$$ solutions $$(x,y,z)$$ on the surface $$xyz-ax-ay-bz=d.$$ where $$a,b,d$$ are positive integers.

I can certainly prove that for a fixed $$z$$ there are finitely many non-negative solutions. But I'm having difficult time proving that it has finitely many solutions, even for some special triples $$a,b,d$$.

## 2 Answers

If $$x=0$$, then $$-ay-bz=d$$ where LHS is non-positive while RHS is positive. So, we have $$x\not=0$$. We have $$y\not=0$$ and $$z\not=0$$ similarly.

Now, we may suppose that $$y\ge x\gt 0$$ and $$z\gt 0$$.

If $$\min(x,y,z)=z$$, then $$1=\frac{a}{yz}+\frac{a}{xz}+\frac{b}{xy}+\frac{d}{xyz}\le \frac{a}{z^2}+\frac{a}{z^2}+\frac{b}{z^2}+\frac{d}{z^3}$$ $$\implies z^3-(2a+b)z-d\le 0$$ The number of $$z$$ satisfying this inequality is finite.

Consindering $$(zx-a)(zy-a)=bz^2+dz+a^2$$ for each $$z$$, we see that the number of $$(x,y)$$ is finite.

If $$\min(x,y,z)=x$$, then, similarly as above, the number of solutions is finite.

Example :

For your example where $$a=9,b=4,d=37$$, if $$\min(x,y,z)=z$$, we have $$z^3-22z-37\le 0\implies z\le 5$$

For each $$z=1,2,\cdots, 5$$, consider $$(zx-9)(zy-9)=4z^2+37z+81$$

For example, for $$z=1$$, we have $$(x-9)(y-9)=2\times 61$$ $$\implies (x-9,y-9)=(1,2\times 61),(2,61)$$ $$\implies (x,y)=(10,131),(11,70)$$ since we already supposed that $$y\ge x$$.

• @mathlove.Nice.Thanks. – Jeff Apr 23 at 17:30

There will always be finitely many (positive) solutions.

Supposing $$a, b, c, d$$ are all positive integers, then the positive integer solutions to

$$xyz = ax + by + cz + d \tag{1}$$

are bounded, for the following reason:

Observe that $$(1)$$ is equivalent to

\begin{align} z & = {ax + by + d \over xy + c} \\ & \le {ax + by + d \over xy} \tag2 \\ & = \frac ay + \frac bx + \frac d{xy} \\ & \le a + b + d \tag3 \end{align}

$$(2)$$ because the new denominator is smaller (since $$c \gt 0$$), and $$(3)$$ similarly because $$x, y \ge 1$$.

By an identical argument you will find $$1 \le x \le b + c + d$$ and $$1 \le y \le a + c + d$$

Let $$M = 3 \cdot \max \{a, b, c, d\}$$ just for simplicity's sake; then all positive integer solutions to $$(1)$$ are bounded within the region $$[1, M] \times [1, M] \times [1, M]$$, within which there are only finitely many ($$M^3$$) integral points

• You may have to do some further work beyond this for the case when one or more of $x, y, z$ are zero, but I suspect that will be a simpler problem/degenerate case – Rob Bland Apr 23 at 16:12
• I believe it should be $z = \frac{ax+by+d}{xy\color{red}{-}c}$ – lisyarus Apr 23 at 16:13
• Whoops, oh dear – Rob Bland Apr 23 at 17:32