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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 $a,b,d$.

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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 $y\ge x$.

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  • $\begingroup$ @mathlove.Nice.Thanks. $\endgroup$ – Jeff Apr 23 at 17:30
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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

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  • $\begingroup$ 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 $\endgroup$ – Rob Bland Apr 23 at 16:12
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    $\begingroup$ I believe it should be $z = \frac{ax+by+d}{xy\color{red}{-}c}$ $\endgroup$ – lisyarus Apr 23 at 16:13
  • $\begingroup$ Whoops, oh dear $\endgroup$ – Rob Bland Apr 23 at 17:32

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