Here are a few functional equations that I made up that I thought that the math SE community would enjoy.
Just so everybody knows, I don't "need help" with these problems - they are simply for the enjoyment of anybody who is interested, and they all have non-constant solutions.
Have fun!
$$f(x+1)^2-4f(x)^2=f(2x+1)+2f(2x)$$ $$g(x)+g(x+1)=x+2$$ $$h(x)+h(x+1)=h\bigg(\sqrt{x^3+x^2-2x+1}\bigg)$$
On
For (2)
derivative both sides twice you have $g''(x)+g''(x+1)=0$ now I guessed that $g''(x)=0$ so $g'(x)$ is constant satisfying $g'(x)+g'(x+1)=1$ so $g'(x)=1/2$ so $g(x)=1/2x+c$ and is satisfying $g(x)+g(x+1)=x+2$ we conclude that $c=0.75$. so $g(x)=1/2x+0.75$ satisfying the equation.
On
For $(1)$, note that
$$(f(x+1)+2f(x))(f(x+1)-2f(x))=f(2x+1)+2f(2x)$$
One subset of solutions requires
$$f(x+1)+2f(x)=0$$
By setting $f(x)=2^xg(\pi x)$, this comes down to
$$g(\pi x+\pi)=-g(\pi x)$$
Which has a general Fourier series solution
$$g(x)=\sum_{n=0}^\infty a_{2n+1}\cos((2n+1)x)+b_{2n+1}\sin((2n+1)x)$$
And thus,
$$f(x)=2^x\sum_{n=0}^\infty a_{2n+1}\cos((2n+1)\pi x)+b_{2n+1}\sin((2n+1)\pi x)$$
For $(2)$, let $g(x)=\frac14+\frac12x+f(\pi x)$ to get
$$f(\pi x)=-f(\pi x+\pi)$$
You could consider a Fourier series as the solution for this, mainly,
$$f(x)=\sum_{n=0}^\infty a_{2n+1}\cos((2n+1)x)+b_{2n+1}\sin((2n+1)x)$$
And thus,
$$g(x)=\frac14+\frac12x+\sum_{n=0}^\infty a_{2n+1}\cos((2n+1)\pi x)+b_{2n+1}\sin((2n+1)\pi x)$$
Solutions to (1) include $f(x) = 0$, $f(x) = -1$, and $f(x) = \dfrac{r^x}{r-2}$ where (if you want real solutions) $r > 0$ and $r \ne 2$.
(2) is an inhomogeneous linear equation, so the general solution is a particular solution plus the general solution of the homogeneous equation. Yanko found a particular solution $g(x) = \frac{x}{2} + \frac{3}{4}$. The homogeneous equation $$ g(x) + g(x+1) = 0 $$ has general solution $$ g(x + n) = (-1)^n g(x), \ n \in \mathbb Z, \; x \in [0,1)$$ with $g$ defined arbitrarily on $[0,1)$. If you want the solution to be continuous, require $\lim_{x \to 1-} g(x) = -g(0)$.