$P(0,y): f(f(y))=g(0)$. $P(x,1):g(x)=g(0)-xf(1)$. $P(x,y):f\equiv c$ therefore $f\equiv a$, $g(x)=-ax+a$ hmm.. maybe wrong? (we can elaborate $f(0)$)

Spectralon wrote: $P(0,y): f(f(y))=g(0)$. $P(x,1):g(x)=g(0)-xf(1)$. $P(x,y):f\equiv c$ therefore $f\equiv a$, $g(x)=-ax+a$ hmm.. maybe wrong? Looks correct to me

parmenides51 wrote: Find all functions $f: R \to R ,g: R \to R$ satisfying the following equality $f(f(x+y))=xf(y)+g(x)$ for all real $x$ and $y$ Let $P(x,y)$ be the assertion $f(f(x+y))=xf(y)+g(x)$ subtracting $P(0,x+1)$ from $P(1,x)$, we get $f(x)=g(0)-g(1)$ constant Plugging $f(x)=c$ in original equation, we get $g(x)=c(1-x)$ And so $\boxed{(f(x),g(x))=(c,c(1-x))\quad\forall x}$ which indeed fits.

A typo? Maybe $\boxed{(f(x),g(x))=(c,c(1-x))\quad\forall x}$

Spectralon wrote: A typo? Maybe $\boxed{(f(x),g(x))=(c,c(1-x))\quad\forall x}$ Yes, indeed. Edited. Thanks

f(f(x+y))=x*f(y)+g(x) P(y,x)=f(f(x+y))=x*f(x)+g(y) so, x*f(y)+g(x)=x*f(x)+g(y) x*f(x)-g(x)=x*f(y)-g(y)=t so we got x*f(x)=g(x)+t, g(x)=x*f(x)-t exchange g(x) to x*f(x)-t at first equation then, you got f(f(x+y))=x(f(x)+f(y))-t P(0,y)=f(f(y))=-t so,-t=f(f(x+y))= x(f(x)+f(y))-t x(f(x)+f(y))=0 for all real x,y means,f(x)+f(y)=0 so f(x)=0 Cause I'm beginner of Aops and math,my sol will be not perfect even at the English please Instruct me