# 2003 Pan African MO Problems/Problem 3

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## Problem

Does there exists a base in which the numbers of the form: $10101, 101010101, 1010101010101,\cdots$ are all prime numbers?

## Solution

Using the definition of base numbers, the number $10101$ in base $b$ can be rewritten as $b^4 + b^2 + 1$, where $b \ge 1$.

The above expression can be factored, so \begin{align*} b^4 + b^2 + 1 &= b^4 + 2b^2 + 1 - b^2 \\ &= (b^2 + 1)^2 - b^2 \\ &= (b^2 + b + 1)(b^2 - b + 1) \end{align*} In order for $(b^2 + b + 1)(b^2 - b + 1)$ to be prime, either $b^2 + b + 1$ or $b^2 - b + 1$ (but not both) must equal $1$. If $b^2 + b + 1 = 1$, then $b = 0$ or $b = -1$, but none of the values of $b$ are valid base numbers. If $b^2 - b + 1 = 1$, then $b = 0$ or $b = 1$. However, neither value are valid bases because $0$ is less than $1$ and the number $10101$ has zeroes for digits (making base $1$ an invalid base).

Therefore, there is no base where $10101, 101010101, 1010101010101,\cdots$ are all prime numbers.