2005 IMO Problems/Problem 4


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


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 $1.$

See Also

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