Problem Overview

Let’s introduce the three problems we will tackle:

1. Determine whether (2^n + 5^n) is a prime number, and also prove that (2^{n-1}) is not prime.
2. Using numbers of the form (4p_1p_2...p_k-1), demonstrate that there are infinitely many primes of the form (4n-1). What complications arise if we attempt a similar proof for primes of the form (4n+1)?
3. Let (a < b) be distinct natural numbers. Prove that in every set of (b) consecutive natural numbers, there exist two distinct numbers whose product is a multiple of (ab). If (a < b < c), must every set of (c) consecutive natural numbers contain three distinct numbers whose product is a multiple of (abc)?

Take some time to think through these problems before we work through them together.

Section 1: Solution to Problem 1

To establish that (2^n + 5^n) is not prime, we can approach the solution in two elegant ways, both interconnected. First, we utilize modular arithmetic:

Notice that:

[

2^n equiv -1 (text{mod } 3) quad text{and} quad 5^n equiv -1 (text{mod } 3)

]

Thus, we can evaluate (2^n + 5^n text{mod } 3):

[

2^n + 5^n equiv (-1)^n + (-1)^n equiv 0 (text{mod } 3)

]

This indicates that (2^n + 5^n) is divisible by 3, meaning it cannot be prime.

We can also apply binomial expansions to reinforce this conclusion.

Next, to show (2^{n-1} - 1) is not prime, we can employ polynomial factorization. Recognizing that (91 = 13 times 7), we can express (2^{n-1} - 1) as ((2^{13})^k - 1). Substituting (x = 2^{13}) gives us a polynomial, (x^k - 1), which we can factorize:

When substituting back for (x), we arrive at:

This demonstrates that (2^{n-1} - 1) has at least two factors, neither of which equals 1, confirming that (2^{n-1} - 1) is not prime.

Section 2: Solution to Problem 2

To prove there are infinitely many primes of the form (4n-1), we start by noting that all primes other than 2 are odd and can be expressed as either (4N + 1) or (4N - 1).

Assume we have a finite list of primes (p_1, p_2, ldots, p_k) of the form (4n - 1). Consider the number (n = 4p_1p_2...p_k - 1).

If (n) is prime, our proof concludes since it is of the form (4N - 1) and not part of the original list. If (n) is not prime, it must have at least one prime factor of the form (4N - 1) as the product of primes of the form (4N + 1) yields a number of the form (4N + 1). Since (n) is (4N - 1), it cannot solely consist of primes of the form (4N + 1), allowing us to find a new prime of the form (4N - 1) not in our initial list. Therefore, there are infinitely many primes of the form (4N - 1).

If we attempted the same reasoning for primes of the form (4N + 1), we encounter issues. The product of two primes of the form (4N - 1) results in a number of the form (4N + 1), allowing (n = 4p_1p_2...p_k + 1) to possibly consist solely of primes of that form.

Section 3: Solution to Problem 3

Let (a < b) be distinct natural numbers. We want to demonstrate that every group of (b) consecutive natural numbers contains two distinct numbers whose product is a multiple of (ab).

Consider a sequence of (b) consecutive numbers, such as 6, 7, 8, 9, 10. With (b=5), there will be exactly one number divisible by 5—in this example, it is 10.

For the second part, the reasoning is similar, but we add an extra consideration. In any set of (c) consecutive natural numbers, one will be divisible by (c), so we can conclude there will indeed be three distinct numbers whose product is a multiple of (abc).

I hope you found these three problems both enjoyable and enlightening. Feel free to share your thoughts or insights on these intriguing challenges!

Video Description: A comprehensive introduction to number theory for beginners, covering fundamental concepts and problems.