$abx^2+bcy^2+acz^2=(xyz)^2+2abc$ has no integral solutions if $a,b,c,x,y,z >1$?

Question

let $a,b,c,x,y,z$ be all pairwise coprime integers . Show that: $$abx^2+bcy^2+acz^2=(xyz)^2+2abc$$ has no integral solutions if $a,b,c,x,y,z >1$. I tried to confirm the results in wolfram but I am totally clueless as to how to prove this. Any hints?

Answer 1

your problem is wrong

  a   b   c      x   y  z
 49  39  25     41  11  4  


Sun Dec 20 13:08:27 PST 2015

Answer 2

Note that for example $(a,b,c,x,y,z) = (1,1,7,1,5,3)$ is a solution to this equation.

So, what do you mean by "non-trivial"?

Answer 3

$$abx^2+bcy^2+acz^2=(xyz)^2+2bca$$

$$x=2n+1$$

$$c=2n^2+2n+1$$

$$a=(2n^2+2n+1)y^2+z^2$$

$$b=4n^2(n+1)^2y^2+(2n^2+2n+1)z^2$$