I would like some help with finding all real valued functions that satisfy this equation:
f(f(y)+xf(x))=y+(f(x))2
I tried the usual substitutions like x=y=0, but my experience with this kind of problem is very limited.
EDIT: I'm an idiot and copied the wrong right side. Updated it now.
Answer
f(f(y)+xf(x))=y+f2(x)
Let x=0, then :
f(f(y))=y+f2(0)
So as y is one-one and onto and invertible, that said that f is onto, so there's no harm supposing that f(x)=0 or x=f−1(0):
f(f(y))=y⟹f(y)=f−1y
Does that give you a hint[f is a one-one onto invertible self-inverse function]. Try f(x)=x or f(x)=−x.
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