Let $f$ and $g$ be functions that we call similar if there exists a function $h$ that is defined for all real numbers, takes each real number exactly once, and for all real $x$, it holds that $h(f(x))=g(h(x))$.

For which parameters $a, b$ will the function $x^{2}+a x+b$ be similar to the function $x^{2}?$
Suppose $h$ is a function that satisfies the conditions, that is, it is defined for all real numbers, takes every real value exactly once, and for every real $x$, the following holds:

$$
h\left(x^{2}+a x+b\right)=(h(x))^{2}
$$

The function $(h(x))^{2}$ takes every positive real number exactly twice, and the value $0$ exactly once. The function

$$
x^{2}+a x+b=\left(x+\frac{a}{2}\right)^{2}+b-\frac{a^{2}}{4}
$$

takes its minimum at $x=-\frac{a}{2}$, and every value greater than the minimum is taken exactly twice. From this, it follows - considering the property of the $h$ function - that for $x=-\frac{a}{2}$, both sides of equation (1) give the value $0$, that is,

$$
h\left(b-\frac{a^{2}}{4}\right)=h\left(-\frac{a}{2}\right)=0
$$

Since the function $h$ takes the value $0$ at exactly one point, it follows that

$$
b-\frac{a^{2}}{4}=\frac{a}{2}
$$

is a necessary condition for the two functions to be similar.

The condition (2) is also sufficient, because, for example, the function $h(x)=x+\frac{a}{2}$ satisfies the conditions of the problem.

István Knébel (Budapest, József A. Gymnasium, 4th grade)

Remarks. 1. It is easily verifiable that the functions

$$
h(x)=\left(x+\frac{a}{2}\right)^{2 k-1}(k=0,1,2, \ldots)
$$

all satisfy the conditions of the problem.

2. The function $h$ that satisfies the conditions also has an inverse function, denoted by $h^{-1}$. The function $h^{-1}$ also satisfies the conditions if the roles of $f$ and $g$ are swapped, that is, from the inequality $h(f(x))=g(h(x))$, it follows that $h^{-1}(g(x))=f\left(h^{-1}(x)\right)$. The roles of the functions $f$ and $g$ are thus symmetric.