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

Let $k$ be a positive integer that satisfies the given condition.

For all primes $p>3$ , by Fermat's Little Theorem, $n^{p-1} \equiv 1\pmod p$ if $n$ and $p$ are relatively prime. This means that $n^{p-3} \equiv \frac{1}{n^2} \pmod p$ . 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 all primes $p>3.$ Since $a_2 = 48 = 2^4 \cdot 3,$ we can conclude that $k$ cannot have any prime divisors. Therefore, our answer is only $1.$

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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