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Question

Find all real functions f from RR satisfying the relation f(x2+yf(x))=xf(x+y).

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Solution

By putting x=0, we get f(yf(0))=0
If f(0)0, then yf(0)takesallrealvalueswhenyvariesoverreallineWegetf(x) =0Supposef(0) = 0Takingy = x,wegetf(x^2- xf(x))= 0forallrealxSupposethereexistsx_0\neq 0inRsuchthatf(x_0) = 0Puttingx = x_0inthegivenrelationwegetf(x^2_0)=x_0f(x_0 + y)forally\in RNowtheleftsideisaconstantandhenceitfollowsfaconstantfunction.Buttheonlyconstantfunctionwhichsatisfiestheequationisidenticallyzerofunction,whichisalreadyobtained.Hencewemayconsiderthecasewheref(x) \neq 0forallx \neq 0.Sincef(x^2+xf(x))= 0,weconcludethatx^2+xf(x) = 0forallx \neq 0.Thisimpliesthatf(x) = xforallx\neq 0.Sincef(0) = 0,weconcludethatf(x) = xforallx \in R.Thuswehavetwofunctions:f(x)= 0andf(x) = xforallx \in R$.

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