Let $f(x+y)=2021^xf(y)+2021^yf(x)$ where $f(x)$ is continuous and $f(1)=1$. Find $f(x)$

Question

Let $f(x+y)=2021^xf(y)+2021^yf(x)$ where $f(x)$ is continuous and $f(1)=1$. Find $f(x)$
When $x=y=0$ we have $f(0)=2021^0f(0)+2021^0f(0)=2f(0)$ $\rightarrow f(0)=0$.
When $x=0, y=y$ we have $f(y)=f(y)$
When $x=x, y=0$ we have $f(x)=f(x)$
When $x=y$ we have $f(2x)=2.2021^xf(x)$
Let's $x=\dfrac{x}{2}$
$$\begin{aligned} f(x)&=2.2021^\frac{x}{2}f(\dfrac{x}{2}) =2^2.2021^\frac{x}{2}2021^\frac{x}{4}f(\dfrac{x}{4})\\&=...\\&=2^{n-1}.2021^{x.(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^n})}f(\dfrac{x}{2^n}) \end{aligned} $$ Take $n \to \infty $ we have $f\left( x \right) = {2^\infty }{.2021^x}.f\left( 0 \right)$. I dont have idea to solve this step. My solution is true? please help me. Many thanks to you guys.

Answer 1

Hint: Divide both sides by $2021^{x+y}$ and let $$g(x)=\frac{f(x)}{2021^x}.$$

Then $$g(x+y)=?$$