14. (15 points) Let
$$
A=x^{4}+2 x^{3}-x^{2}-5 x+34 \text {. }
$$

Find the integer values of \( x \) for which \( A \) is a perfect square.
14. Notice that,
$$
A=\left(x^{2}+x-1\right)^{2}-3(x-11) \text {. }
$$

So, when $x=11$, $A=131^{2}$ is a perfect square.
Next, we prove: there are no other integer $x$ that satisfy the condition.
(1) When $x>11$, we have $A>0$, thus, $A>\left(x^{2}+x-2\right)^{2}$.
Therefore, $\left(x^{2}+x-2\right)^{2}<A<\left(x^{2}+x-1\right)^{2}$.

Let $A=y^{2}(y \in \mathbf{Z})$. Then
$$
\begin{array}{l}
|y|>\left|x^{2}+x-1\right| \\
\Rightarrow|y|-1 \geqslant\left|x^{2}+x-1\right| \\
\Rightarrow y^{2}-2|y|+1 \geqslant\left(x^{2}+x-1\right)^{2} \\
\Rightarrow-3(x-11)-2\left|x^{2}+x-1\right|+1 \geqslant 0 .
\end{array}
$$

Solving this inequality yields the integer values of $x$ as $\pm 3, \pm 2, \pm 1, 0, -4, -5$, but the corresponding $A$ for these values are not perfect squares.
In summary, the integer value of $x$ that makes $A$ a perfect square is 11.