1. Given that $a$ is a root of the equation $x^{2}-5 x+1=0$. Then the unit digit of $a^{4}+a^{-4}$ is $\qquad$ .
ニ、1.7.
Obviously, $a^{-1}$ is another root of the equation. Then $a+a^{-1}=5$.
Thus, $a^{2}+a^{-2}=\left(a+a^{-1}\right)^{2}-2=23$, $a^{4}+a^{-4}=\left(a^{2}+a^{-2}\right)^{2}-2=527$.