1. Find all polynomials $P(x)=a x^{2}+b x+c$ with integer coefficients satisfying

$$
1<P(1)<P(2)<P(3) \quad \text { and at the same time } \quad \frac{P(1) \cdot P(2) \cdot P(3)}{4}=17^{2} \text {. }
$$
1. The equality from the problem is equivalent to the equality $P(1) \cdot P(2) \cdot P(3)=4 \cdot 17^{2}$, so the numbers $P(1), P(2), P(3)$ can only be from the set of divisors of the number $4 \cdot 17^{2}$ greater than 1:

$$
2<4<17<2 \cdot 17<4 \cdot 17<17^{2}<2 \cdot 17^{2}<4 \cdot 17^{2}
$$

If it were $P(1) \geqq 4$, the product $P(1) P(2) P(3)$ would be at least $4 \cdot 17 \cdot(2 \cdot 17)=8 \cdot 17^{2}$, which does not satisfy the problem. Therefore, $P(1)=2$ and then it is necessary that $P(2)=17$, because if $P(2)=4$, the given product $4 \cdot 17^{2}$ would have to be divisible by the number $P(1) P(2)=8$, which is not true, and for $P(2) \geqq 2 \cdot 17$ the product $P(1) P(2) P(3)$ would again be too large. For the third unknown value $P(3)$, we get $P(3)=4 \cdot 17^{2} /(2 \cdot 17)=2 \cdot 17$.

The sought coefficients $a, b, c$ are thus exactly the integers that satisfy the system

$$
\begin{aligned}
& P(1)=a+b+c=2, \\
& P(2)=4 a+2 b+c=17, \\
& P(3)=9 a+3 b+c=34 .
\end{aligned}
$$

By solving this system, we get $a=1, b=12, c=-11$.

Conclusion: The problem is satisfied by the unique polynomial $P(x)=x^{2}+12 x-11$.

For a complete solution, award 6 points. For stating the fact that $P(1), P(2), P(3)$ are from the set of divisors of the number $4 \cdot 17$, award 1 point, for listing all divisors do not award points. For deriving the values that $P(1), P(2), P(3)$ must equal, award 3 points. For setting up (1 point) and solving (1 point) the system, award 2 points (award these points even if the values of $P(1), P(2), P(3)$ are determined incorrectly, as long as the system with them is set up and solved correctly).