As the title implies, I'm looking for triples $(a,b,c)$, where $a,b,c$ are nonzero integers, with $$\frac ab+\frac bc=\frac ca$$
I checked the cases $-100<a,b,c<100$ where $a,b,c\neq 0$ (using Wolfram Mathematica), and found no solutions. Because of this, I conjecture there'd be no solutions.
One idea was using AM-GM, but we barely limit out choice of $a,b,c$ with that. Another idea is to write $ca^2+ab^2=bc^2$ and solving this for, say, $a$ to get $$a=-\frac{b^2}{2c}\pm\frac{1}{2c}\sqrt{b^4+4bc^3}$$ so proving $b^4+4bc^3$ is never a square would be sufficient, but no luck with that either.
Thanks in advance!
Write $x=a/b$,$y=b/c$, $z=c/a$ so get the system $x+y=z$ and $x y z=1$, therefore $x y (x+y)=1$, with $x$, $y$,$z$ rationals. Now write $x=m/q$, $y=n/q$ and we get $$m/q \cdot n/q \cdot (m+n)/q = 1$$ or $$m n (m+n) = q^3$$
We may assume that the numbers $m$, $n$, $m+n$ are pairwise relatively prime, otherwise get rid of the common factor. Therefore, $m$, $n$ and $m+n$ are all cubes, impossible ( Fermat for cubes, proved by Euler).