3. If $k$ is an integer greater than 1, $a$ is a root of $x^{2}-k x+1=0$, for any natural number $n$ greater than 10, the units digit of $a^{2 n}+a^{-2 n}$ is always 7, then the units digit of $k$ is $\qquad$
$3,5,7$
3. 【Analysis and Solution】Obviously, $a \neq 0$, because $a$ is a root of $x^{2}-k x+1=0$, so $a+\frac{1}{a}=k$ ( $\frac{1}{a}$ is the other root).
$$
\begin{array}{l}
a^{2}+a^{-2}=\left(a+\frac{1}{a}\right)^{2}-2=k^{2}-2, \\
a^{2^{2}}-a^{-2^{2}}=\left(k^{2}-2\right)^{2}-2, \\
a^{2^{3}}-a^{-2^{3}}=\left[\left(k^{2}-2\right)^{2}-2\right]^{2}-2, \\
\cdots \cdots \cdot \\
a^{2^{n}}+a^{-2^{n}}=\left(a^{2 n-1}+a^{-2^{n-1}}\right)^{2}-2 .
\end{array}
$$

When $k$ is even, it is easy to see that the unit digit of $a^{2 n}+a^{-2^{n}}$ can only be even; when the unit digit of $k$ is 1, the unit digit of $a^{2^{n}}+$ $a^{-2^{n}}$ is 9; when the unit digit of $k$ is 3, the unit digit of $a^{2 n}+a^{-2 n}$ is 7; when the unit digit of $k$ is 5, for $n>1, a^{2 n}+a^{-2 n}$ the unit digit is also 7; when the unit digit of $k$ is 7, the unit digit of $a^{2^{n}}+a^{-2 n}$ is also 7; when the unit digit of $k$ is 9, the unit digit of $a^{2^{n}}+a^{-2^{n}}$ is 9. Therefore, when $k>1$, for any natural number $n$ greater than 10, to make the unit digit of $a^{2^{n}}$ $+a^{-2^{n}}$ always 7, the unit digit of $k$ is 3 or 5 or 7.