How to Show That All Integers of the Form (2^{2^n} - 1) Are Coprime

In this article, we will explore the mathematical proof of coprimality for the integers of the form (2^{2^n} - 1). This proof will not only demonstrate that these numbers are coprime to each other for different values of n, but it will also lead us to a fascinating conclusion about the infinitude of prime numbers.

Step 1: Proving Coprimality

To show that the integers of the form (2^{2^n} - 1) are coprime, we will denote these integers as Fn (2^{2^n} - 1). We want to demonstrate that for different values of n and m, Fn and Fm are coprime, that is, (text{gcd}(F_n, F_m) 1).

Suppose that d is a common divisor of Fn and Fm. This means:

By the definition of d being a common divisor:

(d mid 2^{2^n} - 1) and (d mid 2^{2^m} - 1)

This implies:

(d mid 2^{2^m} - 2^{2^n} 2^{2^n}(2^{2^{m-n}} - 1))

Without loss of generality, we can assume m ≥ n. Then we can factor (2^{2^m} - 2^{2^n}) as follows:

[begin{aligned}2^{2^m} - 2^{2^n} 2^{2^n}2^{2^{m-n}} - 1 ((2^{2^n} - 1) 1)2^{2^{m-n}} - 1 (2^{2^n} - 1)2^{2^{m-n}} (2^{2^{m-n}} - 1)end{aligned}]Since (d) divides (2^{2^n} - 1), we need to consider the implications of (d) being a common divisor. Given that (d mid 2^{2^n} - 1), (d) can only be a power of 2 (since (2^{2^n}) is even, and (2^{2^n} - 1) is odd). However, since (2^{2^n} - 1) is odd, (d) must be odd. Hence, (d) cannot divide (2^{2^n}), which implies (d) cannot be a power of 2.

We can now deduce that if (d) divides both (2^{2^m} - 1) and (2^{2^n} - 1), then (d) must also divide (2^{2^m} - 2^{2^n}). Using the properties of powers, we can see that:

[ begin{aligned}d mid 2^{2^m} - 2^{2^n} d mid 2^{2^m} - 1 - (2^{2^n} - 1) d mid 2^{2^m} - 2^{2^n}end{aligned} ]Since (d) divides both (2^{2^m} - 1) and (2^{2^n} - 1), it must also divide (2^{2^m} - 2^{2^n}), leading to the conclusion that (d) is odd and must divide both expressions, which is impossible unless (d 1).

Thus, we conclude:

[ text{gcd}(F_n, F_m) 1 quad text{for different } n text{ and } m.

Step 2: Infinitely Many Primes

Now that we have established the coprimality of the integers (2^{2^n} - 1), we can deduce the infinitude of prime numbers. Since (2^{2^n} - 1) is coprime for different values of n, each (2^{2^n} - 1) must be either prime or have distinct prime factors.

If we suppose that there are only finitely many primes, then there would exist some (2^{2^n} - 1) that must share a prime factor with one of the finite set of primes. However, since (2^{2^n} - 1) is coprime to all previously considered forms (2^{2^k} - 1) for (k leq n), it would not be divisible by any of the finite primes, leading to a contradiction. Therefore, we conclude that there must be infinitely many primes.

Conclusion

In conclusion, we have shown that all integers of the form (2^{2^n} - 1) are coprime, which in turn implies the existence of infinitely many primes.

Key Takeaways:

The (2^{2^n} - 1) numbers are coprime for different values of n.

The proof of coprimality leads to the conclusion of the infinitude of prime numbers.

This mathematical exploration highlights the power of number theory and the elegance of proofs in revealing fundamental truths about the nature of prime numbers.