# 2011 USAMO Problems/Problem 4

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 , the remainder upon dividing by is a power of 4. Either prove the assertion or find (with proof) a counter-example.

## 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.

## Solution 2

Lemma (useful for all situations): If and are positive integers such that divides , then divides . Proof: . Replacing the with a and dividing out the powers of two should create an easy induction proof which will be left to the reader as an Exercise.

Consider . We will prove that this case is a counterexample via contradiction.

Because , we will assume there exists a positive integer such that divides and . Dividing the powers of from LHS gives divides . Hence, divides . Because is odd, divides . Euler's theorem gives and so . However, , a contradiction. Thus, is a valid counterexample.
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