1. let $k \geq 0$ be an integer. Determine all real polynomials $P$ of $\operator name{degree} k$ with $k$ different real zeros such that $a$ holds for all zeros of $P$:

$$
P(a+1)=1
$$
Answer: The solutions are all constant polynomials $P(x)=c$ with $c \neq 0$ and all polynomials $P(x)=x+b$ with $b \in \mathbb{R}$.

Solution 1: All constant polynomials $P(x)=c \neq 0$ are solutions. Let $P$ now be such a polynomial with degree at least 1. The polynomial $Q(x)=P(x+1)-P(x)-1$ has by assumption at least $n$ different zeros. However, since $Q$ has a maximum of $\operator name{degree} n-1$, it follows that $Q(x)=0$ or $P(x+1)=P(x)+1$ applies to all $x \in \mathbb{R}$. It follows that $P$ has degree 1, because if this were not the case, a coefficient comparison of $x^{n-1}$ would lead to $a_{n-1}+n a_{n}=a_{n-1}$ (where $\left.P(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{0}\right)$ and thus imply $a_{n}=0$, which of course cannot be the case. So $P$ has degree 1 and we can write $P(x)=a x+b$. Let $y$ be its zero. By assumption

$$
1=P(y+1)=a(y+1)+b=a y+b+a=P(y)+a=a
$$

So $a=1$ must apply. For every $b \in \mathbb{R}$ the polynomial $P(x)=x+b$ is a solution.