2003 Pan African MO Problems/Problem 3

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 ones for digits (making base $1$ an invalid base).

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

2003 Pan African MO (Problems)
Preceded by Problem 2	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 4
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2003 Pan African MO Problems/Problem 3

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