Proving The Inequality 3^a Mod 2^a Is Not Equal To 3^b Mod 2^b For A > B > 2

In the realm of number theory, exploring the relationships between different mathematical expressions often leads to fascinating insights and challenges. This article delves into a specific problem concerning modular arithmetic and exponential functions. The core question we aim to address is whether it's possible to find two distinct integers, $a$ and $b$, both greater than 2, such that $3^a$ modulo $2^a$ is equal to $3^b$ modulo $2^b$. In mathematical notation, we're investigating if the statement $\forall a>b>2 : 3^a \pmod {2^a} \not= 3^b \pmod {2^b}$ holds true. This involves a rigorous examination of the properties of modular arithmetic, exponential functions, and their interplay. To provide a comprehensive understanding, we'll explore the problem's background, the underlying concepts, and a potential approach to proving the given inequality. This exploration will not only solidify our understanding of number theory but also sharpen our problem-solving skills in this domain.

Background and Problem Statement

The heart of our inquiry lies in understanding the behavior of the function $f(n) = 3^n \pmod {2^n}$ for integers $n > 2$. Specifically, we define $r_n = 3^n \pmod {2^n}$, where $0 < r_n < 2^n$. The fundamental question is whether there exist two distinct integers $a$ and $b$, both greater than 2, such that $r_a = r_b$. In simpler terms, we are asking if the remainders of $3^a$ divided by $2^a$ and $3^b$ divided by $2^b$ can ever be the same for $a > b > 2$. This question touches upon the intricate relationships between exponential growth and modular arithmetic. To tackle this, we assume, for the sake of contradiction, that such $a$ and $b$ exist and then proceed to analyze the consequences of this assumption. This approach, known as proof by contradiction, is a powerful tool in mathematical reasoning, allowing us to explore the logical implications of an assumption and potentially arrive at a contradiction, thereby proving the original statement.

Setting up the Proof by Contradiction

To embark on this proof, we assume, without loss of generality (WLOG), that there exist integers $a$ and $b$ such that $a > b > 2$ and $3^a \pmod {2^a} = 3^b \pmod {2^b}$. This assumption is the cornerstone of our proof by contradiction. If we can show that this assumption leads to a logical inconsistency, we can confidently conclude that the original statement, $\forall a>b>2 : 3^a \pmod {2^a} \not= 3^b \pmod {2^b}$, must be true. Given our assumption, we denote the common remainder as $r$, so we have $3^a \equiv r \pmod {2^a}$ and $3^b \equiv r \pmod {2^b}$. This means that there exist integers $k_1$ and $k_2$ such that $3^a = k_1 2^a + r$ and $3^b = k_2 2^b + r$. The challenge now is to manipulate these equations and uncover the implications of this equality in the context of modular arithmetic. This setup allows us to explore the relationships between $3^a$, $3^b$, $2^a$, and $2^b$ under the condition that their remainders upon division are the same.

Exploring the Implications of Equal Remainders

From the equations $3^a = k_1 2^a + r$ and $3^b = k_2 2^b + r$, we can deduce that $3^a - 3^b = k_1 2^a - k_2 2^b$. This equation reveals a crucial relationship between the powers of 3 and powers of 2. Factoring out $3^b$ on the left side, we get $3^b(3^{a-b} - 1) = k_1 2^a - k_2 2^b$. This form highlights that the difference between $3^a$ and $3^b$ is divisible by $3^b$, and it also involves powers of 2. Further manipulation involves factoring out $2^b$ on the right side, leading to $3^b(3^{a-b} - 1) = 2^b(k_1 2^{a-b} - k_2)$. This equation is pivotal because it establishes that $2^b$ divides the left-hand side. Since $3^b$ and $2^b$ are relatively prime (they share no common factors other than 1), it follows that $2^b$ must divide $(3^{a-b} - 1)$. This conclusion is a significant step forward, as it links the difference of a power of 3 and 1 to a power of 2. This connection is what we'll exploit further to potentially arrive at a contradiction.

Delving Deeper into the Divisibility Condition

The fact that $2^b$ divides $(3^{a-b} - 1)$ is a crucial piece of the puzzle. This can be expressed as $3^{a-b} \equiv 1 \pmod {2^b}$. Let's denote $n = a - b$, where $n$ is a positive integer since $a > b$. Now, our condition becomes $3^n \equiv 1 \pmod {2^b}$. This congruence relation suggests that the order of 3 modulo $2^b$ must divide $n$. The order of an integer $a$ modulo $m$ is the smallest positive integer $k$ such that $a^k \equiv 1 \pmod m$. Understanding the order of 3 modulo $2^b$ is essential for making further progress. To determine this order, we can utilize the Lifting The Exponent Lemma (LTE), a powerful tool in number theory for analyzing divisibility properties of exponential expressions. The LTE lemma provides a way to compute the highest power of a prime $p$ that divides $a^n - b^n$ under certain conditions. Applying the LTE lemma in this context can provide insights into the relationship between $n$ and $b$, potentially revealing constraints that lead to a contradiction.

Application of Lifting The Exponent Lemma (LTE)

The Lifting The Exponent Lemma (LTE) is a powerful tool for determining the highest power of a prime dividing a difference of powers. Specifically, LTE can help us find the largest integer $k$ such that $2^k$ divides $3^n - 1$. The LTE lemma states that if $x$ and $y$ are odd integers, and $n$ is an even integer, then $v_2(x^n - y^n) = v_2(x - y) + v_2(x + y) + v_2(n) - 1$, where $v_2(m)$ denotes the exponent of the highest power of 2 that divides $m$. In our case, we have $3^n - 1$, and we want to find $v_2(3^n - 1)$. However, we can't directly apply the LTE lemma as stated because it requires $n$ to be even. To circumvent this, we consider two cases: $n$ is even and $n$ is odd. If $n$ is even, let $n = 2m$. Then $3^n - 1 = 3^{2m} - 1 = (3^m - 1)(3^m + 1)$. We can then apply LTE to each factor separately if needed. If $n$ is odd, we can write $3^n - 1 = (3 - 1)(3^{n-1} + 3^{n-2} + \cdots + 1) = 2(3^{n-1} + 3^{n-2} + \cdots + 1)$. The sum in the parenthesis has $n$ terms, each of which is odd, so the sum is odd. Therefore, if $n$ is odd, $v_2(3^n - 1) = 1$.

Analyzing the Cases and Reaching a Potential Contradiction

Now, let's revisit our condition $3^n \equiv 1 \pmod {2^b}$, which implies that $2^b$ divides $3^n - 1$. Using the LTE Lemma, we can analyze the cases for $n$ being even and odd separately.

Case 1: n is odd. As we derived from the LTE lemma application, if $n$ is odd, then $v_2(3^n - 1) = 1$. This means that the highest power of 2 dividing $3^n - 1$ is $2^1 = 2$. However, we have the condition that $2^b$ divides $3^n - 1$. If $b > 1$, then $2^b$ must be a higher power of 2 than 2. This creates a contradiction, since $b > 2$ by our initial assumptions, so $2^b$ should at least be 8. Therefore, $n$ cannot be odd.

Case 2: n is even. If $n$ is even, let $n = 2m$. Then, $3^n - 1 = 3^{2m} - 1 = (3^m - 1)(3^m + 1)$. Let's examine the divisibility of $(3^m - 1)$ and $(3^m + 1)$ by powers of 2. Notice that $3^m + 1 = (3^m - 1) + 2$. If a power of 2 divides $3^m - 1$, then the next higher power of 2 cannot divide $3^m + 1$, since their difference is only 2. This observation is vital because it limits the combined power of 2 that can divide $3^n - 1$. Applying the LTE lemma in a more refined manner to these factors might lead us to a tighter bound on $v_2(3^n - 1)$. The key here is to understand how the powers of 2 are distributed between the factors $(3^m - 1)$ and $(3^m + 1)$.

The detailed analysis of these cases, especially the even case, requires a further dive into the properties of $v_2(3^m - 1)$ and $v_2(3^m + 1)$. We aim to show that even when $n$ is even, the power of 2 dividing $3^n - 1$ is not high enough to satisfy $2^b$ divides $3^n - 1$ for $b > 2$. This would then complete our proof by contradiction.

Concluding the Proof (To be Completed)

To fully conclude the proof, we need to rigorously analyze Case 2, where $n$ is even. This involves a detailed examination of the factors $(3^m - 1)$ and $(3^m + 1)$ and how powers of 2 divide them. The goal is to demonstrate that even in this case, the condition $2^b$ divides $3^n - 1$ cannot hold for $b > 2$. This would establish the contradiction we initially sought, thus proving that the assumption of the existence of such $a$ and $b$ is false. The final step would be to formally state the conclusion, asserting that for all $a > b > 2$, $3^a \pmod {2^a} \not= 3^b \pmod {2^b}$.

This article has explored a challenging problem in number theory, focusing on the behavior of modular exponentiation. We set out to prove that $\forall a>b>2 : 3^a \pmod {2^a} \not= 3^b \pmod {2^b}$. We established the foundation for a proof by contradiction, delving into the implications of assuming the contrary. We utilized the powerful Lifting The Exponent Lemma to analyze divisibility properties and uncovered constraints that hinted at a contradiction. While the complete conclusion requires further analysis of the even case, the framework and the tools employed here provide a solid pathway to solving this intriguing problem. The journey through this problem showcases the elegance and depth of number theory, highlighting the intricate relationships between seemingly simple mathematical concepts.