Difference between revisions of "2011 USAMO Problems/Problem 4"

Revision as of 10:42, 29 April 2011

Problem

Consider the assertion that for each positive integer $n \ge 2$ , the remainder upon dividing $2^{2^n}$ by $2^n-1$ is a power of 4. Either prove the assertion or find (with proof) a counterexample.

Solution

We will show that $n = 25$ is a counter-example.

Since $2^n \equiv 1 \pmod{2^n - 1}$ , we see that for any integer $k$ , $2^{2^n} \equiv 2^{(2^n - kn)} \pmod{2^n-1}$ . Let $0 \le m < n$ be the residue of $2^n \pmod n$ . Note that since $m < n$ and $n \ge 2$ , necessarily $2^m < 2^n -1$ , and thus the remainder in question is $2^m$ . We want to show that $2^m \pmod {2^n-1}$ is an odd power of $2$ for some $n$ , and thus not a power of $4$ .

Let $n=p^2$ for some odd prime $p$ . Then $\varphi(p^2) = p^2 - p$ . Since $2$ is co-prime to $p^2$ , we have $2^{\varphi(p^2)} \equiv 1 \pmod{p^2}$ and thus $2^{p^2} \equiv 2^{(p^2 - p) + p} \equiv 2^p \pmod{p^2}$ .

Therefore, for a counter-example, it suffices that $2^p \pmod{p^2}$ be odd. Choosing $p=5$ , we have $2^5 = 32 \equiv 7 \pmod{25}$ . Therefore, $2^{25} \equiv 2^7 \pmod{25}$ and thus $2^{2^{25}} \equiv 2^7 \pmod {2^{25} - 1}$ . Since $2^7$ is not a power of $4$ , we are done.

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 ==See also==
+*[[USAMO Problems and Solutions]]
 {{USAMO newbox|year=2011|num-b=3|num-a=5}}

2011 USAMO (Problems • Resources)
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Difference between revisions of "2011 USAMO Problems/Problem 4"

Revision as of 10:42, 29 April 2011

Problem

Solution

See also