$Z=(N-XY)^2$ is a surface with absolute minima ($0s$) anywere $Y=N/X$.

I know this question is naiive, but shouldn't it be possible to apply a [lossy compression](https://i.stack.imgur.com/cg87V.png) filter to this function which preserves values only at discrete boundaries and then search for absolute minima?

If the answer is no, is there any online reference which describes the difficulties of this approach?

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### POSTSCRIPT

Some things crystalised in the discussion below, and need more space than can be added in the comments.

First, this question originates from the interaction of 3D rendering software and its interaction with the function $Z=(N-XY)^2$ (tested for small N only). The interaction results in a surface denoted by $LC(x,y)$ below. 
![enter image description here](https://i.stack.imgur.com/ecs8g.png)

Now let $f(x)=N/x$, and $m(x)=\frac{df}{dx}=-N/x^2$, then the line normal to $f(x)$ at $\mathbf{x^\prime}$ is:

$$(y-f(\mathbf{x^\prime})) + (x-\mathbf{x^\prime})/m(\mathbf{x^\prime}) = 0$$ 

Expressing $y$ as a function of $x$:

$$y(x) = \frac{N^2-\mathbf{x^\prime}^4}{N\mathbf{x^\prime}} + x\frac{\mathbf{x^\prime}^2}{N} $$ 

Let the mouth of the valley below $\{\mathbf{x^\prime},f(\mathbf{x^\prime}),z\}$ be:

$$
r(\mathbf{x^\prime},z)=\sqrt{(x-\mathbf{x^\prime})^2+(y(x)-f(\mathbf{x^\prime}))^2} \,\,\, \text{where} \,\, LC(x,y(x))=z
$$ 

What is the likelihood of finding a suitable function $LC(x,y)$ such that $r(\mathbf{x^\prime},z)$ is maximum when $\mathbf{x^\prime}$ and $f(\mathbf{x^\prime})$ are integers?

Some things to note about $LC(x,y)$ and $r(\mathbf{x^\prime},z)$

* $r(\mathbf{x^\prime},0) = 0$ iff $\{\mathbf{x^\prime},f(\mathbf{x^\prime}),0\}$ is rendered (i.e. sampled); otherwise $r(\mathbf{x^\prime},0)$ is undefined.
* $r(\mathbf{x^\prime},z)$ is defined for $z>=z\_{sat}$ where $z\_{sat}$ is implementation dependent.
* if $LC(x,y)$ is perfectly lossless, then $r(\mathbf{x^\prime},z)$ can be computed for all positive $\mathbf{x^\prime}$ and $z$ by solving the quadratic: $$\sqrt{z}-N+x\frac{N^2-\mathbf{x^\prime}^4}{N\mathbf{x^\prime}} + x^2\frac{\mathbf{x^\prime}^2}{N}=0$$