Let an=(3+√5)^n+(3-√5)^n. The remainder when lfloor(3+√5)^21rfloor is divided by 8 is ____.

Since $3\pm\sqrt5$ are the roots of $t^2-6t+4=0$, the sequence satisfies $a_{n+2}=6a_{n+1}-4a_n$ with $a_0=2$, $a_1=6$, so every $a_n$ is an integer. Because $0<3-\sqrt5<1$, we have $0<(3-\sqrt5)^{21}<1$, hence $(3+\sqrt5)^{21}=a_{21}-(3-\sqrt5)^{21}$ lies strictly between $a_{21}-1$ and $a_{21}$, so its floor is $a_{21}-1$. Reducing the recurrence modulo $8$ gives $a_0,\dots,a_4\equiv 2,6,4,0,0$; since $a_{n+2}\equiv6a_{n+1}-4a_n$, two consecutive zero residues force every later term to be $0$, so $a_n\equiv0\pmod 8$ for all $n\ge3$ (and indeed $a_{21}=1256350449401856$ is divisible by $8$). Thus $\lfloor(3+\sqrt5)^{21}\rfloor\equiv-1\equiv7\pmod 8$.