Difference between revisions of "2011 USAMO Problems/Problem 4"
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We will show that <math>n = 25</math> is a counter-example. | We will show that <math>n = 25</math> is a counter-example. | ||
− | Since <math>2^n \equiv 1 \pmod{2^n - 1}</math>, we see that for any integer <math>k</math>, <math>2^{2^n} \equiv 2^{(2^n - kn)} \pmod{2^n-1}</math>. Let <math>0 \le m < n</math> be the residue of <math>2^n \pmod n</math>. Note that since <math>m < n</math> and <math>n \ge 2</math>, necessarily <math>2^m < 2^n -1</math>, and thus the remainder in question is <math>2^m</math>. We want to show that <math>2^m \pmod {2^n-1}</math> is an odd power of | + | Since <math>\textstyle 2^n \equiv 1 \pmod{2^n - 1}</math>, we see that for any integer <math>k</math>, <math>\textstyle 2^{2^n} \equiv 2^{(2^n - kn)} \pmod{2^n-1}</math>. Let <math>0 \le m < n</math> be the residue of <math>2^n \pmod n</math>. Note that since <math>\textstyle m < n</math> and <math>\textstyle n \ge 2</math>, necessarily <math>\textstyle 2^m < 2^n -1</math>, and thus the remainder in question is <math>\textstyle 2^m</math>. We want to show that <math>\textstyle 2^m \pmod {2^n-1}</math> is an odd power of 2 for some <math>\textstyle n</math>, and thus not a power of 4. |
− | Let <math>n=p^2</math> for some odd prime <math>p</math>. Then <math>\varphi(p^2) = p^2 - p</math>. Since | + | Let <math>\textstyle n=p^2</math> for some odd prime <math>\textstyle p</math>. Then <math>\textstyle \varphi(p^2) = p^2 - p</math>. Since 2 is co-prime to <math>\textstyle p^2</math>, we have |
+ | <cmath>{2^{\varphi(p^2)} \equiv 1 \pmod{p^2}}</cmath> | ||
+ | and thus | ||
+ | <cmath>\textstyle 2^{p^2} \equiv 2^{(p^2 - p) + p} \equiv 2^p \pmod{p^2}.</cmath> | ||
− | Therefore, for a counter-example, it suffices that <math>2^p \pmod{p^2}</math> be odd. Choosing <math>p=5</math>, we have <math>2^5 = 32 \equiv 7 \pmod{25}</math>. Therefore, <math>2^{25} \equiv | + | Therefore, for a counter-example, it suffices that <math>\textstyle 2^p \pmod{p^2}</math> be odd. Choosing <math>\textstyle p=5</math>, we have <math>\textstyle 2^5 = 32 \equiv 7 \pmod{25}</math>. Therefore, <math>\textstyle 2^{25} \equiv 7 \pmod{25}</math> and thus |
+ | <cmath>\textstyle 2^{2^{25}} \equiv 2^7 \pmod {2^{25} - 1}.</cmath> | ||
+ | Since <math>\textstyle 2^7</math> is not a power of 4, we are done. | ||
==See also== | ==See also== |
Revision as of 11:47, 6 May 2011
Problem
Consider the assertion that for each positive integer , the remainder upon dividing by is a power of 4. Either prove the assertion or find (with proof) a counterexample.
Solution
We will show that is a counter-example.
Since , we see that for any integer , . Let be the residue of . Note that since and , necessarily , and thus the remainder in question is . We want to show that is an odd power of 2 for some , and thus not a power of 4.
Let for some odd prime . Then . Since 2 is co-prime to , we have and thus
Therefore, for a counter-example, it suffices that be odd. Choosing , we have . Therefore, and thus Since is not a power of 4, we are done.
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 |