105. Is it true that there exists a number $C$ such that for all integers $k$ the inequality

$$
\left|\frac{k^{8}-2 k+1}{k^{4}-3}\right|<C ?
$$
105. Let's see how the expression $\left|\frac{k^{3}-2 k+1}{k^{4}-3}\right|$ behaves for large (in absolute value) values of $k$. Clearly, in the numerator, the term $k^{3}$ plays the main role, and in the denominator, $k^{4}$. Therefore, we can expect that for large values of $k$, our expression is approximately equal to $\left|\frac{k^{3}}{k^{4}}\right|=\frac{1}{|k|}$. Now let's investigate how much the exact value of our expression differs from the found approximate value. For this, we will perform the following transformation:

$$
\begin{aligned}
&\left|\frac{k^{3}-2 k+1}{k^{4}-3}\right|=\left|\frac{k^{3}\left(1-\frac{2}{k^{2}}+\frac{1}{k^{3}}\right)}{k^{4}\left(1-\frac{3}{k^{4}}\right)}\right|= \\
&=\frac{1}{|k|} \frac{\left|1-\frac{2}{k^{2}}+\frac{1}{k^{3}}\right|}{\left|1-\frac{3}{k^{4}}\right|}
\end{aligned}
$$

Let $|k| \geqslant 2$; then

$$
\begin{gathered}
\left|1-\frac{2}{k^{2}}+\frac{1}{k^{3}}\right| \leqslant 1+\frac{2}{k^{2}}+\frac{1}{|k|^{3}} \leqslant \\
\leqslant 1+\frac{1}{2}+\frac{1}{8}\frac{1}{2}
\end{gathered}
$$

Therefore, for $|k| \geqslant 2$, the inequality

$$
\frac{1}{|k|} \frac{\left|1-\frac{2}{k^{2}}+\frac{1}{k^{3}}\right|}{\left|1-\frac{3}{k^{4}}\right|}<\frac{1}{2} \cdot \frac{2}{\frac{1}{2}}=2
$$

holds. Thus, for $|k| \geqslant 2$, our expression does not exceed 2. It remains to see what values it takes for $k=$ $=-1,0,1$. These values are respectively $1,1 / 3,0$.

Answer: the number $C$ exists, for example, we can take

$$
C=2
$$

Remark. In the same way, we can obtain more accurate estimates for our expression and verify that the maximum value, equal to 1, it takes at $k=-1$. Do this independently.