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Difference between revisions of "2003 Pan African MO Problems/Problem 3"

(Solution to Problem 3 -- easy number base problem)
 
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&= (b^2 + b + 1)(b^2 - b + 1)
 
&= (b^2 + b + 1)(b^2 - b + 1)
 
\end{align*}</cmath>
 
\end{align*}</cmath>
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In order for <math>(b^2 + b + 1)(b^2 - b + 1)</math> to be prime, either <math>b^2 + b + 1</math> or <math>b^2 - b + 1</math> (but not both) must equal <math>1</math>.  If <math>b^2 + b + 1 = 1</math>, then <math>b = 0</math> or <math>b = -1</math>, but none of the values of <math>b</math> are valid base numbers.  If <math>b^2 - b + 1 = 1</math>, then <math>b = 0</math> or <math>b = 1</math>.  However, neither value are valid bases because <math>0</math> is less than <math>1</math> and the number <math>10101</math> has zeroes for digits (making base <math>1</math> an invalid base).
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In order for <math>(b^2 + b + 1)(b^2 - b + 1)</math> to be prime, either <math>b^2 + b + 1</math> or <math>b^2 - b + 1</math> (but not both) must equal <math>1</math>.  If <math>b^2 + b + 1 = 1</math>, then <math>b = 0</math> or <math>b = -1</math>, but none of the values of <math>b</math> are valid base numbers.  If <math>b^2 - b + 1 = 1</math>, then <math>b = 0</math> or <math>b = 1</math>.  However, neither value are valid bases because <math>0</math> is less than <math>1</math> and the number <math>10101</math> has ones for digits (making base <math>1</math> an invalid base).
  
 
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Latest revision as of 03:44, 29 July 2023

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.

See Also

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