Let us consider the function f in three varibles x,y,z $$ f(x,y,z)=(axy+bxz+cyz)/dxyz$$

where $a,b,c,d$ are rational numbers

My question is: Show that there exist still $x,y,z$ positive integers such that $f(x,y,z)$ is also a positive integer.

  • 1
    $\begingroup$ What’s $d$? Any way that would be a neat trick if I set $a=\pi, b=c=1$ $\endgroup$ – Macavity Oct 29 '17 at 7:32
  • $\begingroup$ @Macavity: The question is edited. $\endgroup$ – Safwane Oct 29 '17 at 7:36

This doesn't hold for arbitrary rational numbers $a,b,c,d$. A counter-example is this: consider the case $a=b=c=1$ and $d=10$, then the function can be written as (assuming $x,y,z>0$) $$f(x,y,z)= \frac{1}{10z} + \frac{1}{10y} + \frac{1}{10x},$$ which means that for all positive integers $x,y,z$ we have that $f(x,y,z)\le 3/10$, thus $f(x,y,z)$ cannot be a positive integer for any positive integers $x,y,z$.


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