Problem 28 - Precalculus: Mathematics for Calculus

Question

Show that \(x+y\) is a factor of \(x^{2 n-1}+y^{2 n-1}\) for all natural numbers \(n\)

Step-by-Step Solution

Answer

\(x+y\) is a factor of \(x^{2n-1} + y^{2n-1}\) for all natural numbers \(n\).

Step 1: Understand the Problem

The task is to prove that \(x+y\) is a factor of the polynomial \(x^{2n-1} + y^{2n-1}\) for all natural numbers \(n\). This means for a given polynomial, its division by \(x+y\) should have no remainder.

Step 2: Base Case Verification

Let's first verify this for the smallest natural number \(n=1\). For \(n=1\), the expression is \(x^{2(1)-1} + y^{2(1)-1} = x + y\). Clearly, \(x+y\) is a factor of itself.

Step 3: Use Polynomial Remainder Theorem

According to the Polynomial Remainder Theorem, if \(x+y\) is a factor of \(x^{2n-1} + y^{2n-1}\), then substituting \(x = -y\) in the expression should yield zero. Let's test that.

Step 4: Substitute and Simplify

Substituting \(x = -y\) in \(x^{2n-1} + y^{2n-1}\) gives:\((-y)^{2n-1} + y^{2n-1} = -y^{2n-1} + y^{2n-1}.\)This simplifies to \(0\), indicating \(x + y\) is indeed a factor for any natural \(n\).

Step 5: Generalize for All Natural Numbers

Since the step worked for \(n=1\) and checking with \(x=-y\) makes the expression zero, it stands true that \(x+y\) is a factor for any natural number \(n\). By induction, this can be shown to hold for all other natural numbers.

Key Concepts

Polynomial Remainder Theorem

The Polynomial Remainder Theorem is a vital tool in polynomial algebra. It states that when a polynomial \( f(x) \) is divided by a linear divisor \( x - c \), the remainder of this division is simply \( f(c) \). This means to find out if \( x + y \) is a factor, we can substitute \( x = -y \) in the polynomial.

This theorem is especially useful when proving that a factor exists for all natural numbers \( n \). By applying it to the given polynomial \( x^{2n-1} + y^{2n-1} \), and substituting \( x = -y \), we can check that:

• The expression simplifies to 0 when \( x = -y \).
• This gives a remainder of 0, establishing \( x + y \) as a factor of the polynomial.

The ability to substitute and check values in real time helps confirm divisibility quickly. When \( f(x) = x^{2n-1} + y^{2n-1} \), substituting \( x = -y \) results in \( f(-y) = -y^{2n-1} + y^{2n-1} = 0 \), proving \( x + y \) is a factor.

Factor of a Polynomial

Understanding a polynomial factor is about recognizing expressions that can perfectly divide a polynomial without leaving a remainder. If \( x+y \) is a factor of \( x^{2n-1} + y^{2n-1} \), then dividing the polynomial by \( x+y \) will not leave anything behind.

Factors play a crucial role because they simplify polynomial operations and help solve polynomial equations efficiently. In this case:

• The expression \( x^{2n-1} + y^{2n-1} \) can be thought of as arranged terms that \"cancel out\" each other's powers upon division, which is confirmed using the Polynomial Remainder Theorem.
• Recognizing that \( x + y \) divides it evenly means knowing for certain that when we divide this polynomial by \( x + y \), the remainder is zero, i.e., it fits perfectly, confirming it is indeed a factor.

Thus, identifying \( x + y \) as a factor of the polynomial indicates that this element fits seamlessly into the expression, facilitating various algebraic processes.

Mathematical Induction

Mathematical Induction is a powerful proof technique used to assert the truth of an infinitely large number of propositions. It's akin to the domino effect: by proving a base case, and showing that if one instance holds true, all subsequent ones must follow suit, you demonstrate the truth of a statement for an entire set.

Applying this to our exercise:

• The base case starts with \( n = 1 \), showing \( x+y \) is a factor of \( x + y \) itself—itself is trivial confirmation.
• Then, using the induction hypothesis, assume the expression is valid for an arbitrary \( n \). You must show it continues to hold for \( n + 1 \).
• This approach effectively covers all natural numbers, as validating subsequent cases ensures the polynomial's factorization property persists throughout.

Using Mathematical Induction is key in proofs like these, where an endless number of natural numbers must validate the factorization property. This method not only offers rigor but ensures the result is universally applicable.