4. Find all positive integers $k \leqslant 100$, such that there exists an integer $n$, satisfying $3 n^{6}+26 n^{4}+33 n^{2}+1$ is a multiple of $k$.


Solution
Note: Due to the computational volume, the positive integers in this problem are changed to "composite numbers". Let $f(n)=3 n^{6}+26 n^{4}+33 n^{2}+1$, and consider the prime factors $p \leqslant 31$ of $k$.
(1) From $f(n) \equiv n+n+1 \equiv 1(\bmod 2)$, we know $p \neq 2$;
(2) $f(1)=63=3^{2} \times 7$, so $p=3,7$;
(3) $n^{6}-n^{2}=n^{2}\left(n^{2}+1\right)\left(n^{2}-1\right) \equiv 0(\bmod 5) \Rightarrow f(n) \equiv 3 n^{2}+n^{4}+3 n^{2}+1$ $=\left(n^{2}+3\right)^{2}-8 \equiv 1,3(\bmod 5)$, so $p \neq 5$;
(4) $f(n) \equiv 3 n^{6}+4 n^{4}+1(\bmod 11)$, and the quadratic residues modulo 11 are $0,1,-2,3,4,5$, so $f(n) \equiv 1,-3,3,4$ $(\bmod 11) \Rightarrow p \neq 11$;
(5) $f(2)=741=3 \times 13 \times 19$, so $p=13,19$;
(6) $f(n) \equiv 3 n^{6}+9 n^{4}-n^{2}+1(\bmod 17)$, and the quadratic residues modulo 17 are $0, \pm 1, \pm 2, \pm 4, \pm 8$, so $p \neq 17$; (7) $f(n) \equiv 3 n^{6}+3 n^{4}+10 n^{2}+1(\bmod 23)$, and the quadratic residues modulo 17 are $0,1,2,3,4,-5,6,-7,8,9$, $-10,-11$, so $p \neq 23$;
(8) $f(7)=416991=3 \times 29 \times 4793$, so $p=29$;
(9) $f(12)=9501841=31 \times 306511$, so $p=31$.

Therefore, the prime factors of the composite number $k$ are $3,7,13,19,29,31$.
Hensel's Lemma Corollary: Let $f$ be a polynomial with integer coefficients, $p$ a prime, and $n>1$ an integer. If $a \in \mathbf{Z}$ satisfies $f(a) \equiv 0$ $(\bmod p)$ and $\left(p, f^{\prime}(a)\right)=1$, then the congruence equation $f(x) \equiv 0\left(\bmod p^{n}\right)$ has a unique solution $b$ satisfying $b \equiv a(\bmod p)$. That is, a solution $a$ of the congruence equation $f(x) \equiv 0(\bmod p)$, as long as $p$ does not divide $f^{\prime}(a)$, can be uniquely lifted to a solution of $f(x) \equiv 0\left(\bmod p^{n}\right)$.
From $f^{\prime}(n)=18 n^{5}+104 n^{3}+66 n$, we know $3 \nmid f^{\prime}(1)$, so by the above corollary, the equation has solutions $k=9,27,81$; similarly, $7 \nmid f^{\prime}(1)$, so by the above corollary, the equation has a solution $k=49$;
Also, by the Chinese Remainder Theorem, we know the equation has solutions $k=3 \times 7=21,3 \times 13=39,3 \times 19=57,3 \times 29=87,3 \times 31=93$, $7 \times 9=63,7 \times 13=91$.
In summary, the set of composite numbers $k$ that satisfy the problem and are no more than 100 is $\{9,21,27,39,49,57,63,81,87,91,93\}$.