I want to find all functions $f:\mathbb R\to \mathbb R$ which satisfy $$f(x^4+2x^2+2)-1=(f(x)-1)^4+4x(f(x)-1)^2+4x^2$$ for all $x$.
I know the solution: the only solution is $f(x)=x$. How can I prove it?
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If you know the solution why don't you just plug it in and show the equality of both sides? – AnotherPerson Dec 30 '15 at 00:24
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2@SirJective yes, but how to show that f(x)=x is the only the only solution? – piteer Dec 30 '15 at 08:09
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More hypotheses are needed to ensure a unique solution, perhaps the OP meant to assume continuity.
Call $g(x) = f(x) - 1$. The functional equation is: $$g\circ p(x) = (g(x)^2 + 2x)^2$$ for $p(x) = x^4 + 2x^2 + 2$. Since the image of $p(x)$ is $[2,\infty)$ and the expression containing $g(x)$ on the right hand side is squared, you can safely modify the values of $g$ on the domain $(-\infty, 2)$ by a minus sign while preserving the equation.
Ben
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