Question

Show that if \(2^{n}-1\) is prime, then \(n\) is prime.

Short answer

Answer: Yes, if \(2^{n} - 1\) is prime, then \(n\) must be prime.

Step-by-step solution

Assume \(n\) is not prime

Let's assume the opposite of what we are trying to prove: that \(n\) is not prime. If \(n\) is not prime, then it can be written as the product of two positive integers \(a\) and \(b\), where \(a > 1\) and \(b > 1\). So we have \(n = ab\).

Rewrite \(2^{n}\) using \(2^{ab}\)

Now let's rewrite \(2^{n}\) using the product \(ab\), so we have \(2^{n} = 2^{ab}\).

Use properties of exponents to rewrite \(2^{ab}\)

We can now use properties of exponents to rewrite \(2^{ab}\) as \((2^{a})^{b}\).

Apply the difference of squares factorization

Recall that \(2^{n} - 1\) is prime. Since \(2^{n} = (2^{a})^{b}\), we can write this prime number as \(2^{n} - 1 = (2^{a})^{b} - 1\). Now notice the structure of this expression as a difference of squares: $$(2^{a})^{b} - 1 = ((2^{a})^{b} - 1)((2^{a})^{b} + 1) = (2^{ab} - 1)(2^{ab} + 1).$$

Show that \(2^{n} - 1\) is not prime

From the above equation, we can now see that \(2^{n} - 1 = (2^{ab} - 1)(2^{ab} + 1)\). Since \(a > 1\) and \(b > 1\), we know that \(2^{ab} - 1 > 1\) and \(2^{ab} + 1 > 1\). Thus, we have found two factors of \(2^{n} - 1\), which implies that \(2^{n} - 1\) is not prime.

Reach a contradiction

However, our assumption was that \(2^{n} - 1\) is prime. Since we have found a contradiction in our assumption, we can now conclude that our initial assumption (that \(n\) is not prime) is false. So, if \(2^{n} - 1\) is prime, then \(n\) must be prime.

Key concepts

Integer Factorization

Integer factorization is the process of breaking down a composite number into smaller integers, called factors, which multiply together to recreate the original number. In simpler terms, it's like reverse multiplication. Let's consider 15 as an example. The integer factorization of 15 is 3 and 5, because when you multiply them you achieve the original number: 3 × 5 = 15.

Factorization becomes especially significant when dealing with prime numbers. A prime number is an integer greater than one that has no positive divisors other than 1 and itself. When checking if a particular number is prime, often a factorization attempt is made first.

In our exercise, this concept comes into play with the assumption that if \( n \) is not prime, it can be expressed as the product of two numbers, \( a \) and \( b \). If we express this product, they should both be greater than one, thus showing that \( n \) was not originally prime. The contradiction arises when factoring leads to an implication that revealed \( 2^n - 1 \) was not prime when it was assumed to be.

Exponent Properties

Exponent properties, or rules, are essential tools in simplifying mathematical expressions when dealing with powers. Here’s a quick rundown of some key points:

• When multiplying, add the exponents: \( a^n \times a^m = a^{n+m} \)
• When raising a power to another power, multiply the exponents: \( (a^n)^m = a^{n \times m} \)
• Any number to the power of zero is one: \( a^0 = 1 \) (given \( a eq 0 \))

In the exercise, the exponent property used allows us to transition from \( 2^{ab} \) to \( (2^a)^b \,\) showcasing the fundamental capability to rearrange and simplify powers in different scenarios. This skill is essential to understand because if you mishandle exponents, it might lead to conclusions that aren't true! So, remembering these basic rules can make working with exponential equations like \( 2^n - 1\) in the exercise more manageable.

Difference of Squares

The difference of squares is a special algebraic formula that helps factor certain types of expressions to simplify or solve them. This formula is \( a^2 - b^2 = (a - b)(a + b) \), where \( a \) and \( b \) are any expressions. It is derived from the multiplication of binomials and is used when you have two terms being subtracted from one another.

In our exercise, when \( 2^n - 1 \) gets rewritten using the properties of exponents, it bears resemblance to the difference of squares structure. While \( 2^n \) is not literally a square, the underlying method of refactoring still uses this principle to express \( 2^{ab} - 1 \) in a manageable way. We essentially treat part of the expression like a squared term for the purposes of the factorization argument.

This aspect of the solution instrumentalizes the factorization, resting on the idea that differing squares always yield factors. Thus, recognizing this pattern in expressions like \( 2^n - 1 \) can simplify complex problems and thereby uncover deeper mathematical truths like the one implied in the exercise.