For how many integers $x$ will the following polynomial be a perfect square?

$$
4 x^{4}-12 x^{3}+17 x^{2}-6 x-14
$$
Let $x_{0}$ be a fixed integer, and we are looking for square numbers close to $p\left(x_{0}\right)$. For this purpose, we seek an integer-coefficient polynomial whose square is close to $p(x)$. It is easy to find one such that the first two coefficients of the square match the first two coefficients of $p(x)$:

$$
q_{c}(x)=\left(2 x^{2}-3 x+c\right)^{2}=4 x^{4}-12 x^{3}+(9+4 c) x^{2}-6 c x+c^{2}
$$

and with this, the difference is:

$$
\delta_{c}(x)=p(x)-q_{c}(x)=4(2-c) x^{2}+6(c-1) x-\left(4+c^{2}\right)
$$

The difference $\delta_{2}(x)$ corresponding to the coefficient $c=2$ is linear, while the differences corresponding to the neighboring integer $c$ values are quadratic polynomials:

$$
\begin{array}{ll}
c=1 \text { to }: & \delta_{1}(x)=4 x^{2}-15 \\
c=3 \text { to }: & \delta_{3}(x)=-4 x^{2}+12 x-23
\end{array}
$$

The latter is negative for all $x$, $\delta_{1}(x)$ is positive if $|x|>\sqrt{15} / 2$. In this case,

$$
q_{1}(x)<p(x)<q_{3}(x)
$$

If therefore $|x| \geq 2$, then $p\left(x_{0}\right)$ falls between the squares of the integers $2 x_{0}^{2}-3 x_{0}+1$ and $2 x_{0}^{2}-3 x_{0}+3$, so it can only be a square if it equals the square of $2 x_{0}^{2}-3 x_{0}+2$, that is, if $\delta_{2}\left(x_{0}\right)=0$, from which $x_{0}=3$, and thus $p(3)=11^{2}$.

The values of $p(x)$ for the remaining $\left|x_{0}\right| \leq 1$ to be examined are:

$$
p(-1)=25, \quad p(0)=-14, \quad p(1)=-11
$$

Among these, only the first is a square number.

Accordingly, $p(x)$ produces a perfect square for exactly two integer $x$ values. smallskip

János Reviczky (Budapest, I. István Gymnasium)

It was also solved by József Beck, Péter Maróti, Viktor Martoni, Gábor Somorjai.

As a prize, József Beck and Viktor Martoni received a 50 Ft book voucher.