101. The number $N=2^{4 n+1}-4^{n}-1$ is divisible by 9 for an integer $n \geqslant 0$. The number $8 N+9$ is a perfect square.

10
101. $N=2^{4 n+1}-4^{n}-1=2 \cdot 4^{2 n}-4^{n}-1=\left(2 \cdot 4^{n}+\right.$ $+1)\left(4^{n}-1\right)=\left(2^{2 n+1}+1\right)\left(4^{n}-1\right)$, both factors are multiples of 3, therefore, $N$ is a multiple of 9. In addition, $8 N+9=$ $=\left(4^{n+1}-1\right)^{2}$.