Question

Find a factor of \(2^{1001}-1\).

Short answer

Answer: A non-trivial factor of \(2^{1001}-1\) is \(2^{500}-1\).

Step-by-step solution

Write the expression using difference of squares

Rewrite \(2^{1001}-1\) as \((2^{500})^2 -1^2\).

Factor using difference of squares

Using the difference of squares fact \((a^2-b^2)=(a+b)(a-b)\), we can factor the expression: \((2^{500})^2 -1^2 = (2^{500}+1)(2^{500}-1)\).

Write the expression using difference of cubes

Next, rewrite \(2^{1001}-1\) as \((2^{333})^3 -1^3\).

Factor using difference of cubes

Using the difference of cubes fact \((a^3-b^3)=(a-b)(a^2+ab+b^2)\), we can factor the expression: \((2^{333})^3 -1^3 = (2^{333}-1)((2^{333})^2+2^{333}+1)\).

Identify a non-trivial factor

We can see that \(2^{500}-1\) is a factor which is not trivial, therefore: A non-trivial factor of \(2^{1001}-1\) is \(2^{500}-1\).

Key concepts

Difference of Squares

The difference of squares is a classic mathematical concept used in factorization. It can be expressed as \((a^2-b^2)\), which simplifies to \((a+b)(a-b)\). This is a very useful formula because it makes factoring large powers manageable. For example, consider \(2^{1001}-1\). This expression can be seen as \((2^{500})^2 - 1^2\). By treating it as a difference of squares, you can factor it into \((2^{500}+1)(2^{500}-1)\). The process transforms a complicated expression into simpler terms, which are easier to handle.

Difference of Cubes

The difference of cubes is another useful pattern in factorization. Unlike squares, cubes involve three-dimensional terms. The formula for the difference of cubes is \((a^3-b^3) = (a-b)(a^2 + ab + b^2)\). This allows a seemingly complex expression to be simplified. In the case of \(2^{1001}-1\), it was rewritten as \((2^{333})^3 - 1^3\) to fit this pattern. This simplifies to \((2^{333}-1)((2^{333})^2+2^{333}+1)\), breaking down the expression into easily factorable components. This approach is invaluable when dealing with high powers in expressions.

Non-Trivial Factor

A non-trivial factor is any factor of a number that is not 1 or the number itself. When solving equations or problems involving factorization, finding non-trivial factors is crucial. In the given exercise, a non-trivial factor of \(2^{1001}-1\) is \(2^{500}-1\). This implies that besides 1 and \(2^{1001}-1\), \(2^{500}-1\) divides the expression leaving a remainder of zero, making it an essential part of the factorization. Recognizing and extracting non-trivial factors helps in simplifying and solving complex mathematical expressions effectively.

Number Theory

Number theory is a vast and vital branch of mathematics dealing with the properties and relationships of numbers, especially integers. It encompasses everything from elementary conceptions like divisibility to more complicated ideas such as primality and factorization. In the context of the exercise \(2^{1001}-1\), number theory provides the foundation for understanding why methods, like the difference of squares or cubes, work. It enables mathematicians to discover non-trivial factors, like \(2^{500}-1\), by applying theoretical concepts. Understanding the fundamental principles in number theory enhances one's ability to tackle diverse mathematical problems involving integers and their factors. Number theory is critical in cryptography, computer science, and advanced mathematics, making it an integral part of mathematical studies.