contest-mathfunctional-equations
Find all $f(x)$ satisfying $f(f(x)) = x^2 – 2$.
Presumably $f(x)$ is supposed to be a function from $\mathbb R$ to $\mathbb R$ with no further restrictions (we don't assume continuity, etc), but the text of the problem does not specify further.
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Source: This question. It is about to be closed for containing too many problems in one question. I'm posting each problem separately.
It is not true that $|ax|=a|x|$; the correct identity is $|ax|=|a||x|$.
Whether or not adding the hypothesis of continuity is necessary for additive functions to be linear depends on the axiom of choice. Using a Hamel basis $B$ for $\mathbb{R}^n$ over $\mathbb{Q}$ together with one of its countable subsets $A=\{x_1,x_2,\ldots\}$, you can construct a discontinuous $\mathbb{Q}$ linear map from $\mathbb{R}^n$ to $\mathbb{R}$ by taking the unique $\mathbb{Q}$ linear extension of the function $f:B\to\mathbb{R}$ such that $f(x_k)=k|x_k|$ and $f(B\setminus A)=\{0\}$. Since $\mathbb{R}$ linear maps between finite dimensional real vector spaces are continuous, such a map cannot be linear. However, it is consistent with ZF that all additive functions from $\mathbb{R}^n$ to $\mathbb{R}$ are continuous (I am however not knowledgeable in the set theoretic background needed to show this).
Solution: Taking it as known that the perpendicular bisectors of a triangle are concurrent, and that this implies the concurrence of the altitudes of a triangle, our construction and proof are simple.
Draw lines from the endpoints $E_1$ and $E_2$ of the circle through the point (call it $A$), meeting the circle in $B$ and $C$; now extend $E_1C$ and $E_2B$ until they meet in $D$. Now, by Thales' theorem, $E_1B$ and $E_2C$ are altitudes of $\triangle E_1DE_2$, meeting in the point $A$. Thus, the third altitude of this triangle, dropped from $D$, also passes through $A$; that is, $DA$ is the third altitude of the triangle. As such, it is perpendicular to the diameter of the circle.
Best Answer
There is no such $f$.
From http://yaroslavvb.com/papers/rice-when.pdf , the question of existence is determined by the theorem:
Theorem 6. Let $\mathbb{R}$ be the real line. Let $g$ be a real quadratic polynomial, so that $$g(x)=ax^2+ (b + 1)x+c,$$ for all real $x$, where $a\ne 0$, $b$, and $c$ are in $\mathbb{R}$. ... set $\Delta(g)= b^2-4ac$. If $\Delta(g)> 1$, then g has no iterative roots of any order whatever. [That is, there is no $f$ such $f\circ f = g$.] If $\Delta(g) =1$, then $g$ can be embedded in a 2-sided flow on $\mathbb{R}$, all of whose members are continuous functions. If $\Delta(g) <1$, then $g$ can be embedded in a 1-sided flow on $\mathbb{R}$, all of whose members are continuous functions; but $g$ cannot be embedded in any 2-sided flow on $\mathbb{R}$.
As $\Delta(g) = 0 - 4(1)(-2) = 8 > 1$ in your case, the question of existence is negative.
Looking closely at the article, the main point is that no function with only one 2-cycle can have a square root. In our case that means that there can be no partial solution $f:D\to D$ of the funcional equation $f(f(x))=x^2-2$ in $D\subset\Bbb{R}$ if $x_0=\frac{-1+\sqrt{5}}{2}\approx 0.61803$ or $x_2=\frac{-1-\sqrt{5}}{2}\approx -1.61803$ are in $D$.
In fact, clearly $x_0^2-2=x_2$ and $x_2^2-2=x_0$ (this implies that $x_0\in D$ if and only if $x_2\in D$).
There can be no other pair $y_1\ne y_2$ with $y_1^2-2=y_2$ and $y_2^2-2=y_1$, since then $$ \{-1,2,x_0,x_2,y_1,y_2\} $$ would be roots of the polynomial $P(x)=x^4 - 4 x^2 - x + 2$, since $y_1^2-2=y_2$ and $y_2^2-2=y_1$ implies $$ (y_1^2-2)^2-2=y_1\quad\Rightarrow \quad y_1^4-4y_1^2+2=y_1 \quad\Rightarrow \quad P(y_1)=0 $$ and similarly $P(y_2)=0$.
Now, if $x_0\in D$ (or $x_2\in D$) and $f:D\to D$ satisfy $f(f(x))=x^2-2$, then $x_1:=f(x_0)$ and $x_3:=f(x_2)$ would be such a pair, a contradiction that proves $x_0\notin D$ (and $x_2\notin D$).