Question

Find all real functions $f$ from $R\to R$ satisfying the relation $f(x^2+yf(x\left)\right)=xf(x+y)$.

Answer 1

By putting $x=0$, we get $f\left(yf\right(0\left)\right)=0$
If $f\left(0\right)\ne 0$, then $yf(0$)$takesallrealvalueswhen$y$variesoverreallineWeget$f(x) =0$Suppose$f(0) = 0$Taking$y = x$,weget$f(x^2- xf(x))= 0$forallrealxSupposethereexists$x_0\neq 0$inRsuchthat$f(x_0) = 0$Putting$x = x_0$inthegivenrelationweget$f(x^2_0)=x_0f(x_0 + y)$forall$y\in R$Nowtheleftsideisaconstantandhenceitfollows$f$a^{\prime }constantfunctio{n}^{\prime }.Buttheonlyconstantfunctionwhichsatisfiestheequationisidenticallyzerofunction,whichisalreadyobtained.Hencewemayconsiderthecasewhere$f(x) \neq 0$forall$x \neq 0$.Since$f(x^2+xf(x))= 0$,weconcludethat$x^2+xf(x) = 0$forall$x \neq 0$.Thisimpliesthat$f(x) = x$forall$x\neq 0$.Since$f(0) = 0$,weconcludethat$f(x) = x$forall$x \in R$.Thuswehavetwofunctions:$f(x)= 0$and$f(x) = x$forall$x \in R$.