For the polynomial $f$, it holds that $f\left(x^{2}+1\right)-f\left(x^{2}-1\right)=4 x^{2}+6$. Determine the polynomial $f\left(x^{2}+1\right)-f\left(x^{2}\right)$.
Since $f$ is a non-constant polynomial, otherwise

$$
f\left(x^{2}+1\right)-f\left(x^{2}-1\right)=c-c=0
$$

If $f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$, then

$$
f\left(x^{2}+1\right)=a_{n}\left(x^{2}+1\right)^{n}+a_{n-1}\left(x^{2}+1\right)^{n-1}+\cdots+a_{1}\left(x^{2}+1\right)+a_{0}
$$

and

$$
f\left(x^{2}-1\right)=a_{n}\left(x^{2}-1\right)^{n}+a_{n-1}\left(x^{2}-1\right)^{n-1}+\cdots+a_{1}\left(x^{2}-1\right)+a_{0} .
$$

Consider the general term of the difference of the two polynomials:

$$
d_{k}(x)=a_{k}\left[\left(x^{2}+1\right)^{k}-\left(x^{2}-1\right)^{k}\right]
$$

Using the identity $u^{k}-v^{k}=(u-v)\left(u^{k-1}+u^{k-2} v+\cdots+v^{k-1}\right)$, we get that $d_{k}(x)=2 a_{k}\left[\left(x^{2}+1\right)^{k-1}+\right.$ $\left.\left(x^{2}+1\right)^{k-2}\left(x^{2}-1\right)+\cdots+\left(x^{2}-1\right)^{k}\right]$. The expression in the brackets is a sum of polynomials of degree $(2 k-2)$, each with a leading coefficient of 1, so $d_{k}(x)$ is either the zero polynomial - if $a_{k}=0$ - or precisely of degree $(2 k-2)$.

Among the non-zero terms of the difference $f\left(x^{2}+1\right)-f\left(x^{2}-1\right)$, there are no equal degrees, so the degree of the difference is the degree of the highest degree term, which is $2 n-2$.

According to the condition, the polynomial $f\left(x^{2}+1\right)-f\left(x^{2}-1\right)$ is quadratic, so it is precisely of degree $(2 n-2)$ and thus $n=2$. In this case, $f(x)=a_{2} x^{2}+a_{1} x+a_{0}$, and the difference $f\left(x^{2}+1\right)-f\left(x^{2}-1\right)$ is

$$
a_{2}\left(x^{2}+1\right)^{2}+a_{1}\left(x^{2}+1\right)+a_{0}-a_{2}\left(x^{2}+1\right)-a_{1}\left(x^{2}+1\right)-a_{0}=4 a_{2} x^{2}+2 a_{1} .
$$

According to the condition, this difference is $4 x^{2}+6$. Two polynomials are equal if and only if the coefficients of the terms of the same degree are equal, from which $a_{2}=1$ and $a_{1}=3$. Thus, the polynomial $f(x)$ is of the form $x^{2}+3 x+a_{0}$, where $a_{0}$ can be any real number. Therefore,

$$
f\left(x^{2}+1\right)-f\left(x^{2}\right)=\left(x^{2}+1\right)^{2}+3\left(x^{2}+1\right)+a_{0}-\left[\left(x^{2}\right)^{2}+3 x^{2}+a_{0}\right]=2 x^{2}+4
$$

András Kôrizs (Budapest, Veres Péter Gymnasium, 12th grade)