parmenides51 wrote: Find all functions $f : R \to R$ defined for real numbers, take real values and for all real $x$ and $y$ the equality holds: $$f(2^x+2y)=2^yf(f(x))f(y).$$ $\boxed{\text{S1 : }f(x)=0\quad\forall x}$ is a solution. So let us from now look only for not allzero solutions. Let $P(x,y)$ be the assertion $f(2^x+2y)=2^yf(f(x))f(y)$ Let $a=f(0)$ Let $u$ such that $f(u)=v\ne 0$ If $f(x_1)=f(x_2)$, comparaison of $P(x_1,y)$ and $P(x_2,y)$ implies $f(2y+2^{x_1})=f(2y+2^{x_2})$ and so $f(x)$ is periodic. Let $T$ be a period. $P(0,\frac{u-1}2)$ $\implies$ $2^{\frac{u-1}2}f(a)f(\frac{u-1}2)=v\ne 0$ $P(0,\frac{u-1}2+T)$ $\implies$ $2^{\frac{u-1}2}2^Tf(a)f(\frac{u-1}2)=v\ne 0$ And so $2^T=1$ (since $v\ne 0$) and $T=0$. So $f(x)$ is injective. If $f(t)=0$ for some $t$, then $P(x,t)$ $\implies$ $f(2^x+2t)=0$ $\forall x$, impossible since injective So $f(x)\ne 0$ $\forall x$ $P(x,-2^x)$ $\implies$ $f(f(x))=2^{2^x}$ (since $f(x)\ne 0$ $\forall x$) Then $P(x,0)$ $\implies$ $f(2^x)=af(f(x))=a2^{2^x}$ and so $f(x)=a2^x$ $\forall x>0$ (still true when $x=0$) Let any $y$ and $x$ great eough to have $2^x+2y>0$ $P(x,y)$ becomes $a2^{2^x+2y}=2^y2^{2^x}f(y)$ and so $f(y)=a2^y$ $\forall y$ Plugging this back in original equation, we get $a=1$ and $\boxed{\text{S2 : }f(x)=2^x\quad\forall x}$, which indeed fits.