Problem 27 - Algebra and Trigonometry

Question

Show that \(x-y\) is a factor of \(x^{n}-y^{n}\) for all natural numbers \(n\) . \(\left[\text { Hint: } x^{k+1}-y^{k+1}=x^{k}(x-y)+\left(x^{k}-y^{k}\right) y .\right]\)

Step-by-Step Solution

Answer

Using mathematical induction, \( x-y \) is a factor of \( x^n-y^n \) for all natural numbers \( n \).

Step 1: Base Case Verification

For the base case, let \( n = 1 \). We have \( x^1 - y^1 = x - y \). Clearly, \( x - y \) is a factor of itself. Thus, the base case holds true.

Step 2: Inductive Hypothesis

Assume that \( x - y \) is a factor of \( x^k - y^k \) for some natural number \( k \). This means there exists a polynomial \( Q(x, y) \) such that \( x^k - y^k = Q(x, y)(x-y) \).

Step 3: Inductive Step

We need to show that \( x - y \) is also a factor of \( x^{k+1} - y^{k+1} \). Using the provided hint, express:\[x^{k+1} - y^{k+1} = x^k(x-y) + (x^k - y^k)y\]Substitute the inductive hypothesis in:\[x^{k+1} - y^{k+1} = x^k(x-y) + Q(x,y)(x-y)y = (x^k + Q(x,y)y)(x-y)\]Since \( (x^k + Q(x, y)y) \) is a polynomial, \( x - y \) is a factor of \( x^{k+1} - y^{k+1} \).

Step 4: Conclusion

By mathematical induction, since the base case holds and the inductive step shows that if \( x-y \) is a factor of \( x^k - y^k \), it is also a factor of \( x^{k+1} - y^{k+1} \), we conclude that \( x - y \) is a factor of \( x^n - y^n \) for all natural numbers \( n \).

Key Concepts

Factor Theorem

The Factor Theorem is a simple but powerful concept in algebra, especially when dealing with polynomials. At its core, the theorem states that if a polynomial \( f(x) \) has a factor \( x - c \), then \( f(c) = 0 \). This means that the remainder when \( f(x) \) is divided by \( x-c \) is zero.

Understanding this theorem can help in factoring polynomials and solving polynomial equations. If you can find a number \( c \) such that \( f(c) = 0 \), you've found a root and a factor of the polynomial. For example, to determine if \( x-y \) is a factor of a polynomial like \( x^{n} - y^{n} \), you can substitute \( y \) into the polynomial in place of \( x \). If the result is zero, then \( x-y \) is indeed a factor.

In the given exercise, \( x-y \) being a factor means each time \( y \) replaces \( x \), the entire expression equals zero, verifying the factorization.

Polynomials

Polynomials are mathematical expressions involving a sum of powers in one or more variables multiplied by coefficients. They can have different types, such as linear, quadratic, cubic, and so on, based on their degree, which is determined by the highest power of the variable.

A polynomial in one variable (say \( x \)) has the general form:

• \( a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 \)

where \( n \) is a non-negative integer and \( a_n, a_{n-1},..., a_0 \) are constants with \( a_n eq 0 \). The degree of the polynomial is \( n \).

In the exercise at hand, the polynomial \( x^n - y^n \) suggests two variables, making it a bidimensional polynomial when considered in terms of \( x \) and \( y \). This polynomial's distinct property, when considering the factor \( x-y \), allows the use of algebraic manipulation to reveal its complexity in the factorization process.

Inductive Hypothesis

The Inductive Hypothesis is an essential element of the mathematical induction process, a technique used to prove statements, typically about natural numbers. This method involves two primary steps: the base case and the inductive step.

The inductive hypothesis part of the process assumes that a given statement is true for some natural number \( k \). It forms the foundation for proving the next case, \( k+1 \). When tackling our exercise, the statement assumed was that \( x-y \) is a factor of \( x^k - y^k \). The assumption means we start by saying some equation holds for \( k \) and use this to show it holds for \( k+1 \) as well.

This hypothesis is just the middle step in a longer chain to prove the general statement. After establishing that this holds for the base case, and making the inductive assumption for \( k \), we perform algebraic operations to confirm it remains true for \( k+1 \). This systematic approach ensures the original statement's accuracy for all natural numbers \( n \), completing the induction.