Note that this is equivalent to $c=a+\frac{b^2-a}{ab} \implies b|a, a|b^2$. So set $a=bn, b^2=am \implies b^2=bnm \implies b=nm, a=n^2m$. Thus we have $c=\frac{n^3m^2+m-1}{nm} \implies m|1 \implies m=1$ so $c=\frac{n^3}{n}=n^2$.