Solution.
Let $b$ be the base. Then
\begin{eqnarray}
10201 &=& b^4 + 2b^2 + 1
= (b^2+1)^2 \\
10101 &=& b^4 + b^2 + 1 = (b^2+b+1)(b^2-b+1)
\end{eqnarray}
and since $x\ge 2$, both factors exceed 1. Finally
100011 = b^5 + b + 1 = (b^2+b+1)(b^3 -
b^2 + 1)
and again both factors exceed 1.