1. If the two roots of the equation $x^{2}-3 x+1=0$ are also roots of the equation $x^{4}-p x^{2}+q=0$, then the units digit of $(p+q)^{2008}$ is ( ).
(A)2
(B) 4
(C) 6
(D) 8
-1 1.C.
Let the two roots of the equation $x^{2}-3 x+1=0$ be $\alpha, \beta$, then $\alpha^{4}-p \alpha^{2}+q=0, \beta^{4}-p \beta^{2}+q=0$. Also, $\Delta=(-3)^{2}-4=5>0$, so $\alpha \neq \beta$, which means $\alpha^{2} \neq \beta^{2}$. From equation (1), we get
$$
p=\frac{\alpha^{4}-\beta^{4}}{\alpha^{2}-\beta^{2}}=\alpha^{2}+\beta^{2}, q=\frac{\alpha^{2} \beta^{2}\left(\alpha^{2}-\beta^{2}\right)}{\alpha^{2}-\beta^{2}}=\alpha^{2} \beta^{2} \text {. }
$$

Also, $\alpha+\beta=3, \alpha \beta=1$, thus, $\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta=7, \alpha^{2} \beta^{2}=(\alpha \beta)^{2}=1$.
Therefore, $p=7, q=1$.
Thus, $(p+q)^{2008}=8^{2008}=\left(8^{4}\right)^{502}=4096^{x 2}$.