Let $\mathrm{f}\left(\mathrm{x}\right)$ be a polynomial with positive degree satisfying the relation $\mathrm{f}\left(\mathrm{x}\right)\mathrm{f}\left(\mathrm{y}\right)=\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{y}\right)+\mathrm{f}\left(\mathrm{xy}\right)-2$ and $\mathrm{f}\left(4\right)=65$ then

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)\mathrm{f}\left(\mathrm{y}\right)=\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{y}\right)+\mathrm{f}\left(\mathrm{xy}\right)-2 \\ x=y=1\Rightarrow f\left(1\right)=2 \\ y=\frac{1}{x}\Rightarrow f\left(x\right)f\left(\frac{1}{x}\right)=\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\frac{1}{x}\right)\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=1+{\mathrm{x}}^{\mathrm{n}} \end{gathered}$$

$$\begin{gathered} f\left(4\right)=1+4^n=65 \\ {4}^n=64, n=3 \\ f\left(x\right)=1+x^3 \\ \therefore f\left(2\right)=1+2^3=9,f\left(5\right)=1+5^3=126 \\ f\left(x\right)=2x^2 \\ \Rightarrow 1+x^3=2x^2 \\ \Rightarrow x^3-2x^2+1=0 \\ \Rightarrow \left(x-1\right)\left(x^2-x-1\right)=0 \\ x=1,\frac{1\pm \sqrt{5}}{2} \end{gathered}$$