62 Let $a$ be a positive integer, and the sequence $\left\{u_{n}\right\}$ is defined as follows:
$$
u_{1}=2, u_{2}=a^{2}+2, u_{n}=a u_{n-2}-u_{n-1}(n \geqslant 3)
$$

To ensure that the sequence $\left\{u_{n}\right\}$ does not contain any multiples of 4, what conditions should $a$ satisfy?
62 Let $k$ be an integer, and discuss several cases for $a$:
i) $a=4 k$. In this case, $u_{1}$ and $u_{2}$ both leave a remainder of 2 when divided by 4, and $u_{n}=a u_{n-1}-u_{n-2}(n \geqslant 3)$. Using mathematical induction, it can be proven that for any natural number $n$, $u_{n}$ leaves a remainder of 2 when divided by 4. Therefore, in this case, $\left\{u_{n}\right\}$ does not contain any multiples of 4.
ii) $a=4 k+1$. In this case, $u_{2}$ leaves a remainder of 3 when divided by 4. Let $u_{2}=4 p+3$, then $u_{3}=(4 k+1) \cdot 2- (4 p+3)=4(2 k-p-1)+3=4 q+3$, where $p, q$ are both integers, thus $u_{4}=(4 k+1)(4 p+3)-(4 q+3)=4(4 k p+3 k+p-q)$ is a multiple of 4.
iii) $a=4 k+2$. In this case, it is the same as (i), and it can be proven that $u_{n}$ leaves a remainder of 2 when divided by 4.
iv) $a=4 k+3$, in this case, we can set $u_{2}=4 p+3, u_{3}=4 q+3$, thus $u_{4}=(4 k+3)(4 p+3)-(4 q+3)=4 h+2$
$$
u_{5}=(4 k+3)(4 q+3)-(4 h+2)=4 t+3
$$

From this, we know that $u_{3 k+1}$ leaves a remainder of 2 when divided by 4, and the other $u_{n}$ leave a remainder of 3 when divided by 4.
In summary, $a \neq 4 k+1$.