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

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Consider <math>n = 25</math>. We will prove that this case is a counterexample via contradiction.
 
Consider <math>n = 25</math>. We will prove that this case is a counterexample via contradiction.
  
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Because <math>4 = 2^2</math>, we will assume there exists a positive integer <math>k</math> such that <math>2^{2^n} - 2^{2k}</math> divides <math>2^n - 1</math> and <math>2^{2k} < 2^n - 1</math>. Dividing the powers of <math>2</math> from LHS gives <math>2^{2^n - 2k} - 1</math> divides <math>2^n - 1</math>. Hence, <math>2^n - 2k</math> divides <math>n</math>. Because <math>n = 25</math> is odd, <math>2^{24} - k</math> divides <math>25</math>. Euler's theorem gives <math>2^{24} \equiv 2^4 \equiv 16 \pmod{25}</math> and so <math>k \ge 16</math>. However, <math>2^{2k} >= 2^{32} > 2^{25} - 1</math>, a contradiction. Thus, <math>n = 25</math> is a valid counterexample.
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Because <math>4 = 2^2</math>, we will assume there exists a positive integer <math>k</math> such that <math>2^{2^n} - 2^{2k}</math> divides <math>2^n - 1</math> and <math>2^{2k} < 2^n - 1</math>. Dividing the powers of <math>2</math> from LHS gives <math>2^{2^n - 2k} - 1</math> divides <math>2^n - 1</math>. Hence, <math>2^n - 2k</math> divides <math>n</math>. Because <math>n = 25</math> is odd, <math>2^{24} - k</math> divides <math>25</math>. Euler's theorem gives <math>2^{24} \equiv 2^4 \equiv 16 \pmod{25}</math> and so <math>k \ge 16</math>. However, <math>2^{2k} \geq 2^{32} > 2^{25} - 1</math>, a contradiction. Thus, <math>n = 25</math> is a valid counterexample.
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:36, 20 February 2019

This problem is from both the 2011 USAJMO and the 2011 USAMO, so both problems redirect here.

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 counter-example.

Solution

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

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

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

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

Solution 2

Lemma (useful for all situations): If $x$ and $y$ are positive integers such that $2^x - 1$ divides $2^y - 1$, then $x$ divides $y$. Proof: $2^y \equiv 1 \pmod{2^x - 1}$. Replacing the $1$ with a $2^x$ and dividing out the powers of two should create an easy induction proof which will be left to the reader as an Exercise.

Consider $n = 25$. We will prove that this case is a counterexample via contradiction.

Because $4 = 2^2$, we will assume there exists a positive integer $k$ such that $2^{2^n} - 2^{2k}$ divides $2^n - 1$ and $2^{2k} < 2^n - 1$. Dividing the powers of $2$ from LHS gives $2^{2^n - 2k} - 1$ divides $2^n - 1$. Hence, $2^n - 2k$ divides $n$. Because $n = 25$ is odd, $2^{24} - k$ divides $25$. Euler's theorem gives $2^{24} \equiv 2^4 \equiv 16 \pmod{25}$ and so $k \ge 16$. However, $2^{2k} \geq 2^{32} > 2^{25} - 1$, a contradiction. Thus, $n = 25$ is a valid counterexample.

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

See also

2011 USAMO (Problems • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions
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