I would like to show if the following equation is soluble for every positive integer $k$ $$ (5x-1)yz=(2088k+2010+x)(z-y), $$ where $x,y,z$ are positive integers and $z>y$. I am trying to divide $k$ to some congruence classes modulo some number and continue this to cover all cases, but this method takes long time.
My question is first if this equation is soluble for every $k>0$. second if there is a shorter way than investigating residue classes modulo some number?
Thanks.
For $k=1$ the solutions are
$(x=2\land y=82\land z=100)\lor (x=2\land y=400\land z=3280)\lor (x=2\land y=410\land z=4100)\lor (x=2\land y=450\land z=36900)\lor (x=2\land y=451\land z=45100)\lor (x=2\land y=455\land z=373100)\lor (x=6\land y=108\land z=456)\lor (x=7\land y=120\land z=19704)\lor (x=25\land y=21\land z=57)\lor (x=25\land y=33\land z=4389)\lor (x=39\land y=21\land z=1379)\lor (x=87\land y=9\land z=135)\lor (x=273\land y=3\land z=47)\lor (x=1017\land y=1\land z=165)$
Now we know that $y z \geq z-y$ when $z>y>0$, So the above equation could make sense to look for solution when $2088k+2010+x \geq 5x-1$, solving for $x$ we get that $x\leq \frac{1}{4} (2088 k+2011)$, which imply that For any fixed $k$ there are only finitely many solutions.
The rest is to check them by computer search.