Let's find all polynomials $P$ for which the identity

$$
P(x+1)=P(x)+2x+1
$$

holds.
I. solution. Let $P$ be an $n$-degree polynomial, $P(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$, where $a_{n} \neq 0$. It is easy to verify that the polynomial

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

is the sum of at least $(n-1)$-degree terms

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

which means this polynomial is of degree $(n-1)$. In our case, the polynomial $P(x+1)-P(x)=2 x-1$ is of the first degree, so $P$ can only be a quadratic polynomial.

Let $P(x)=a x^{2}+b x+c$. Then

$$
P(x+1)-P(x)=\left(a(x+1)^{2}+b(x+1)+c\right)-\left(a x^{2}+b x+c\right)=2 a x+(a+b)
$$

This is exactly equal to the polynomial $2 x+1$ if $2 a=2$ and $a+b=1$, i.e., $a=1$ and $b=0$. Therefore, the polynomial $P$ can be of the form $x^{2}+c$. It is easy to verify that such polynomials satisfy (1).

II. solution. It is easy to show that polynomials of the form $x^{2}+c$ satisfy (1). We will show that $P$ can only be of this form.

Suppose that the polynomial $P$ satisfies the identity (1), and let $Q(x)=P(x)-x^{2}-P(0)$. For this polynomial, it is true that $Q(0)=P(0)-0^{2}-P(0)=0$, and

$$
Q(x+1)=P(x+1)-(x+1)^{2}-P(0)=P(x)+2 x+1-(x+1)^{2}-P(0)=P(x)-x^{2}-P(0)=Q(x)
$$

for all $x$.

From these two pieces of information, it follows that $Q(x)=0$ for all integer $x$. However, the value of a polynomial can only be zero at infinitely many points if the polynomial is the zero polynomial, i.e., $Q(x)=0$. Thus, $P(x)=x^{2}+P(0)$, meaning $P$ is of the desired form.

Peter Frenkel (Fazekas M. Fóv. Gyak. Gimn., III o.t.)

Remark. Some assumed without justification that the polynomial $P$ can only be quadratic. They received 1 point for their solution.