2005 IMO Problems/Problem 4

Problem

Determine all positive integers relatively prime to all the terms of the infinite sequence $a_n=2^n+3^n+6^n -1,\ n\geq 1.$

Solution

For all primes $p$ greater than $3$ , by Fermat's last theorem, $n^{p-1} = 1$ mod $p$ if $n$ and $p$ are relatively prime. This means that $n^{p-3} = \frac{1}{n^2}$ mod $p$ . Plugging $n = p-3$ back into the equation, we see that the value mod $p$ is simply $\frac{2}{9} + \frac{3}{4} + \frac{1}{36} - 1 = 0$ . Thus, the expression is divisible by $p$ . Because the expression is clearly never divisible by $2$ or $3$ , our answer is all numbers of the form $2^a3^b$ .

2005 IMO (Problems) • Resources
Preceded by Problem 3	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 5
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2005 IMO Problems/Problem 4

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