Logarithm and a special ratio

$\large \log_{32}{x}=\log_{\sqrt{2016}}{y}=\log_{63}{(x+y)}$

$\begin{cases} \dfrac{\log{x}}{\log{32}} = \dfrac{\log{(x+y)}}{\log{63}} \\ \dfrac{\log{y}}{\frac{1}{2}(\log{32}+\log{63})} = \dfrac{\log{(x+y)}}{\log{63}} \end{cases}$

$\Rightarrow \begin{cases} \dfrac{\log{x}}{\log{(x+y)}} = \dfrac{\log{32}}{\log{63}} \\ \dfrac{\log{y^2}}{\log{(x+y)}} = 1+\dfrac{\log{32}}{\log{63}} \end{cases}$

$\dfrac{\log{y^2}}{\log{(x+y)}} = 1+\dfrac{\log{x}}{\log{(x+y)}}$

$\dfrac{\log{y^2}}{\log{(x+y)}} = \dfrac{ \log{(x+y)} + \log{x} }{\log{(x+y)}}$

$\log{y^2}=\log{(x^2+xy)}$

$y^2-xy-x^2=0$

$\left(\dfrac{y}{x}\right)^2-\dfrac{y}{x}-1 =0$

$\dfrac{y}{x}=\dfrac{1+\sqrt{5}}{2}$ or $\dfrac{y}{x}=\dfrac{1-\sqrt{5}}{2}$ (rej.)

$\Rightarrow a+b+c=1+5+2=8$

let z = \log_{32}{x} = \log_\sqrt{2016}{y} = log_{63}{(x+y)}

$32^{z} = x, 2016^{\frac{z}{2}} = y, 63^{z} = x+y \\ 2^{5z} = x, (2^{5}3^{2}7)^{\frac{z}{2}} = y, (3^{2}7)^{z} = x+y$

square the second equation to get $2^{5z}3^{2z}7^{z} = y^{2}$

substitute the first and third equation to get

$x(x+y) = y^{2} \\ (\frac{y}{x})^2-\frac{y}{x}-1=0 \\ \frac{y}{x} = \frac{1\pm\sqrt{5}}{2}$

reject the negative value because this implies one of the first two expressions would the logarithm of negative number which is undefined

therefore $\frac{y}{x} = \frac{1+\sqrt{5}}{2} \Rightarrow a+b+c=1+5+2=8$