Proving the Divisibility of (2^n - 1) by 3

The problem of proving that (2^n - 1) is divisible by 3 for any integer (n) is a classic example of how modular arithmetic can be used to simplify and solve such mathematical problems. In this article, we will explore this problem through a detailed analysis of the expression modulo 3, considering two primary cases based on the parity of (n).

Case 1: (n) is Even

First, let's consider the case where (n) is even. An even number can be written as (n 2k) for some integer (k). Thus, we have:

[2^n 2^{2k} (2^2)^k 4^k]

Next, we need to find the remainder when 4 is divided by 3:

[4 equiv 1 (text{mod} 3)]

Since (4 equiv 1 (text{mod} 3)), we can replace 4 with 1 in the expression (4^k), yielding:

[4^k equiv 1^k (text{mod} 3)]

Therefore:

[2^n - 1 equiv 1 - 1 (text{mod} 3)]

And:

[2^n - 1 equiv 0 (text{mod} 3)]

However, we observe that (2^n - 1) is congruent to 2 modulo 3, which means it is not divisible by 3:

[2^n - 1 equiv 2 (text{mod} 3)]

Hence, when (n) is even, (2^n - 1) is not divisible by 3.

Case 2: (n) is Odd

Now, let's consider the case where (n) is odd. An odd number can be written as (n 2k 1) for some integer (k). Thus, we have:

[2^n 2^{2k 1} 2 cdot 4^k]

Again, since (4 equiv 1 (text{mod} 3)), we can replace 4 with 1 in the expression (4^k), yielding:

[4^k equiv 1^k (text{mod} 3) Rightarrow 4^k equiv 1 (text{mod} 3)]

Therefore:

[2^n equiv 2 cdot 1 (text{mod} 3) Rightarrow 2^n equiv 2 (text{mod} 3)]

When we subtract 1 from both sides, we get:

[2^n - 1 equiv 2 - 1 (text{mod} 3) Rightarrow 2^n - 1 equiv 0 (text{mod} 3)]

This shows that (2^n - 1) is divisible by 3 when (n) is odd.

Conclusion

From the above analysis, we can conclude:

If (n) is even, then (2^n - 1 equiv 2 (text{mod} 3)), and hence it is not divisible by 3. If (n) is odd, then (2^n - 1 equiv 0 (text{mod} 3)), and hence it is divisible by 3.

Additional Analysis: Proving that (n^2 - 1) is Never Divisible by 3

Another related problem is to prove that (n^2 - 1) is never divisible by 3 for any integer (n). We can use modular arithmetic to demonstrate this:

When (n) is divided by 3, it can leave a remainder of 0, 1, or 2. Let's consider the three cases:

Case I: (n 3k)

If (n 3k), then:

[n^2 - 1 (3k)^2 - 1 9k^2 - 1 3(3k^2) - 1]

Since 3(3k^2) is divisible by 3, subtracting 1 from it makes the expression not divisible by 3:

[n^2 - 1 equiv -1 (text{mod} 3) Rightarrow n^2 - 1 equiv 2 (text{mod} 3)]

Case II: (n 3k - 1)

If (n 3k - 1), then:

[n^2 - 1 (3k - 1)^2 - 1 9k^2 - 6k 1 - 1 9k^2 - 6k 3(3k^2 - 2k)]

Since 3(3k^2 - 2k) is divisible by 3, subtracting 1 from it makes the expression not divisible by 3:

[n^2 - 1 equiv -1 (text{mod} 3) Rightarrow n^2 - 1 equiv 2 (text{mod} 3)]

Case III: (n 3k - 2)

If (n 3k - 2), then:

[n^2 - 1 (3k - 2)^2 - 1 9k^2 - 12k 4 - 1 9k^2 - 12k 3 3(3k^2 - 4k 1)]

Since 3(3k^2 - 4k 1) is divisible by 3, subtracting 1 from it makes the expression not divisible by 3:

[n^2 - 1 equiv -1 (text{mod} 3) Rightarrow n^2 - 1 equiv 2 (text{mod} 3)]

Therefore, (n^2 - 1) is never divisible by 3 for any integer (n).