The OP probably meant that $n$ should be odd, otherwise there are too many trivial solutions, e.g. $(x,y,z,n)=(3k,3k,3k,4k)$ for all $k$.

JoeBlow wrote: The OP probably meant that $n$ should be odd I don't see why this would make the problem any harder, take $(x,y,z,n) = (k, 6k, 6k, 3k)$ for odd $k$.

Also, if you look at the rest of the problems from the Kosovo TST, they're all pretty much trivial, so I wouldn't be surprised if this one was too.

$(x,y,z,n)=(n,2n,2n,2n)$ also works. Follow up question(FUQ): Are there infinitely many solutions such that gcd of any 2 variables is 1?

BISHAL_DEB wrote: $(x,y,z,n)=(n,2n,2n,2n)$ also works. Follow up question(FUQ): Are there infinitely many solutions such that gcd of any 2 variables is 1? Obviously not. The equation is equivalent to $x,y,z,n> 0$ and $(x+y)nz=(4z-n)xy$ $x|LHS$ and all variables pairwaise coprime $\implies$ $x=1$ $y|LHS$ and all variables pairwaise coprime $\implies$ $y=1$ $z|RHS$ and all variables pairwaise coprime $\implies$ $z=1$ And then equation is $n>1$ and $2n=4-n$ And so no solution.

modularmarc101 wrote:

How to see that $\frac{1}{3k} + \frac{1}{2k} + \frac{1}{6k} = \frac{4}{4k}$

Please do not revive old topics that have been left alone for 7+ years. $\frac{1}{3k} + \frac{1}{2k} + \frac{1}{6k} = \frac{2}{6k} + \frac{3}{6k} + \frac{1}{6k} = \frac{2+3+1}{6k} = \frac{6}{6k} = \frac{1}{k} = \frac{4}{4k}$

ErijonHasi1 wrote: modularmarc101 wrote:

How to see that $\frac{1}{3k} + \frac{1}{2k} + \frac{1}{6k} = \frac{4}{4k}$ Motivation comes from the known "identity" 1/2 + 1/3 + 1/6 = 1