Find $f(x)$ for which: $f(x^4+2x^2+2)-1=(f(x)-1)^4+4x(f(x)-1)^2+4x^2$

Question

Find $f(x)$ for which: $f(x^4+2x^2+2)-1=(f(x)-1)^4+4x(f(x)-1)^2+4x^2$

I know the solution is $f(x)=x$ but how to prove it

Answer 1

The equation can be written a bit nicer in the following form: $$\tag1f( (x^2+1)^2+1) = ((f(x)-1)^2+2x)^2+1.$$

Note that the the sequence $x_0=0$, $x_{n+1}= x_n^4+2x_n^2+2$ is strictly increasing to $+\infty$. Then the intervals $I_n=[x_n,x_{n+1})$ are a partition of $[0,\infty)$ and $x\mapsto (x^2+1)^2+1$ is a bijection $I_n\to I_{n+1}$.

Let $f_0\colon I_0\to\Bbb R$ be an arbitrary function and recursively define $f_{n+1}\colon I_{n+1}\to\Bbb R$ by letting $f_{n+1}(x)=((f_n(\xi)-1)^2+2\xi)^2+1$ where $\xi=\sqrt{\sqrt{ x-1}-1}$ is the unique element of $I_n$ with $(\xi^2+1)^2+1=x$. Then the $f_n$ together define a function $[0,\infty)\to\Bbb R$ satisfying the functional equation. Using $$ ((f(-x)-1)^2-2x)^2=f((x^2+1)^2+1)=((f(x)-1)^2+2x)^2,$$ i.e., $$\tag2 f(-x) =\begin{cases}\pm\sqrt{(f(x)-1)^2+4x}+1&\text{or alternatively}\\ 1&\text{if }f(x)=1\end{cases} $$ this is readily extended to the negatives.

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In case you are only interested in continuous $f$, make sure that

• $f_0$ is continuous,
• $\lim_{x\to 2^-}f_0(x)=(f_0(0)-1)^4+1$
• always pick the same sign in $(2)$, more specifically, pick "$+$" if $f_0(0)>1$ and pick "$-$" if $f_0(0)<1$.

Answer 2

For convenience, let us rewrite with $g(x):=f(x)-1$,

$$g(x^4+2x^2+2)=(g^2(x)+2x)^2.$$

As $x^4+2x^2+1\ge2$, there is no explicit constraint on $g(x)$ for $x<2$ anf $g(x)$ is free in that range.

Anyway, $x^4+2x^2+1$ is an even function, monotonic in the positives, so that $x^4+2x^2+1$ is also achieved by $-x$ (and no other value). This means that

$$(g^2(x)+2x)^2=(g^2(-x)-2x)^2$$ must hold, i.e.

$$g^2(-x)=g^2(x)+4x$$ or $$g^2(-x)=-g^2(x).$$

The second equation has the useless solution $g(x)=0$ in the reals, and we keep

$$g(-x)=\pm\sqrt{g^2(x)+4x}.$$

Let us restrict to the range $[0,2[$, and let $g(x)=\phi(x)$ for some arbitrary $\phi$.

Then in $[2,26[$,

$$g(x)=(\phi^2(\psi(x))+2\psi(x))^2,$$

where $\psi(x)$ denotes the only positive solution of $\psi^4+2\psi^2+2=x$,

$$\psi(x)=\sqrt{\sqrt{x-1}-1}.$$

(As you can verify, $g(x):=x-1$ in $[0,2[$ does imply $g(x)=x-1$ in $[2,26[$).

One more iteration defines $g(x)$ in the range $[26,458330[$,

$$g(x)=((\phi^2(\psi(x))+2\psi(x))^4+2\psi(x))^2$$

and so on. (The next range is $[458330,44127887745906175987802[$, which should be enough for "practical applications".)

In conclusion, there are uncountably many more solutions than $f(x)=x$. (Not counting the fact that $f(-x)$ can freely alternate in sign.)