1991 USAMO Problems/Problem 3

Problem

Show that, for any fixed integer $\,n \geq 1,\,$ the sequence \[2, \; 2^2, \; 2^{2^2}, \; 2^{2^{2^2}}, \ldots  \pmod{n}\] is eventually constant.

[The tower of exponents is defined by $a_1 = 2, \; a_{i+1} = 2^{a_i}$. Also $a_i \pmod{n}$ means the remainder which results from dividing $\,a_i\,$ by $\,n$.]

Solution

Suppose that the problem statement is false for some integer $n \ge 1$. Then there is a least $n$, which we call $b$, for which the statement is false.

Since all integers are equivalent mod 1, $b\neq 1$.

Note that for all integers $b$, the sequence $2^0, 2^1, 2^2, \dotsc$ is eventually becomes cyclic mod $b$. Let $k$ be the period of this cycle. Since there are $k-1$ nonzero residues mod $b$. $1 \le k\le b-1 < b$. Since \[2, 2^2, 2^{2^2}, 2^{2^{2^2}}, \dotsc\] does not become constant mod $b$, it follows the sequence of exponents of these terms, i.e., the sequence \[1, 2, 2^2, 2^{2^{2}}, \dotsc\] does not become constant mod $k$. Then the problem statement is false for $n=k$. Since $k<b$, this is a contradiction. Therefore the problem statement is true. $\blacksquare$

Note that we may replace 2 with any other positive integer, and both the problem and this solution remain valid.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

1991 USAMO (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5
All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS