7. find all polynomials $Q(x)=a x^{2}+b x+c$ with integer coefficients such that three different primes $p_{1}, p_{2}, p_{3}$ exist with

$$
\left|Q\left(p_{1}\right)\right|=\left|Q\left(p_{2}\right)\right|=\left|Q\left(p_{3}\right)\right|=11
$$
## Solution:

We distinguish between 2 cases, depending on how many signs of $Q\left(p_{k}\right)$ are negative. We can assume that at most one sign is negative, otherwise we replace $Q(x)$ with $-Q(x)$.

(a) $Q\left(p_{1}\right)=Q\left(p_{2}\right)=Q\left(p_{3}\right)=11$

The polynomial $Q(x)-11$ has $\operator name{degree} \leq 2$ and 3 different zeros, so it vanishes identically, therefore $Q(x)=11$ is constant.
(b) $Q\left(p_{1}\right)=-11, Q\left(p_{2}\right)=Q\left(p_{3}\right)=11$

The polynomial $Q(x)-11$ has the zeros $p_{2}, p_{3}$, so we can write $Q(x)-11=a\left(x-p_{2}\right)\left(x-p_{3}\right)$. Inserting $p_{1}$ provides

$$
a\left(p_{1}-p_{2}\right)\left(p_{1}-p_{3}\right)=-2 \cdot 11
$$

If $p_{1}, p_{2}, p_{3}$ are all odd, then the left-hand side is divisible by 4, contradiction. If $p_{1}=2$, then the two brackets on the left are odd and cannot both have a value of 1. A comparison with the right-hand side yields $\left\{p_{2}, p_{3}\right\}=$ $\{3,13\}$. If you insert these values into $Q$ and solve the three equations for the coefficients, the result is $a=-2, b=32, c=-67$. If OBdA $p_{2}=2$, then the first bracket on the left is odd and the second is even. This results in $p_{1}=$ $3, p_{3}=5$ or $p_{1}=13, p_{3}=11$. Analogous calculations as before lead in the first case to $a=11, b=-77, c=121$ and in the second case to $a=-1, b=13, c=-11$.

Overall, we therefore obtain the following possibilities for $\pm Q(x)$

$$
\begin{array}{rll}
11 & ,\left\{p_{1}, p_{2}, p_{3}\right\} \quad \text { any } \\
x^{2}-13 x+11 & ,\left\{p_{1}, p_{2}, p_{3}\right\}=\{2,11,13\} \\
2 x^{2}-32 x+67 & ,\left\{p_{1}, p_{2}, p_{3}\right\}=\{2,3,13\} \\
11 x^{2}-77 x+121 & ,\left\{p_{1}, p_{2}, p_{3}\right\}=\{2,3,5\}
\end{array}
$$