Question

Prove each statement. $$ \begin{aligned} &a-b \text { is a factor of } a^{n}-b^{n}\\\ &\text { [Hint: } \left.a^{k+1}-b^{k+1}=a\left(a^{k}-b^{k}\right)+b^{k}(a-b)\right] \end{aligned} $$

Short answer

By mathematical induction, \( a - b \) is a factor of \( a^n - b^n \).

Step-by-step solution

Understand the Factor Theorem

The Factor Theorem states that if a polynomial \( f(x) \) has a factor \( x-c \), then \( f(c) = 0 \).

Form the Polynomial

Consider the polynomial \( P(a, b) = a^n - b^n \). To prove that \( a - b \) is a factor, we need to show that \( P(a, b) \) vanishes (i.e., equals 0) when \( a = b \).

Use the Hint

By applying the hint provided: \( a^{k+1} - b^{k+1} = a(a^k - b^k) + b^k(a - b) \), let's proceed with induction to prove the statement.

Base Case

For \( n = 1 \), \( a^1 - b^1 = a - b \). Thus, \( a - b \) is clearly a factor.

Induction Hypothesis

Assume for some integer \( k \), we have \( a - b \) is a factor of \( a^k - b^k \), i.e., \( a^k - b^k = (a - b)Q(a, b) \).

Inductive Step

By the given hint and the induction hypothesis: \[ a^{k+1} - b^{k+1} = a(a^k - b^k) + b^k(a - b) \]. Substituting the induction hypothesis into this, we get: \[ a(a^k - b^k) + b^k(a - b) = a((a - b)Q(a, b)) + b^k(a - b) \]. Factoring out \( a - b \), we obtain: \[ a^{k+1} - b^{k+1} = (a - b)(aQ(a, b) + b^k) \].

Conclusion

Since \( a^{k+1} - b^{k+1} \) can be written as \( (a - b)(aQ(a, b) + b^k) \), it follows that \( a - b \) is a factor of \( a^{k+1} - b^{k+1} \). Therefore, by induction, \( a - b \) is a factor of \( a^n - b^n \) for any positive integer \( n \).

Key concepts

Factor Theorem

The Factor Theorem is a key principle in algebra that helps in determining whether a given polynomial has a specific linear factor. It states that for any polynomial \( f(x) \), if \( x = c \) is a root, then \( x - c \) is a factor of \( f(x) \). Here, a root means that \( f(c) = 0 \). This is useful for breaking down complex polynomials into simpler components. For example, to check if \( x - 3 \) is a factor of \( f(x) \), compute \( f(3) \). If \( f(3) = 0 \), then indeed \( x - 3 \) is a factor.

Mathematical Induction

Mathematical induction is a powerful proof technique used to establish the truth of an infinite sequence of statements. It consists of two main steps: the base case and the induction step. In the base case, you show the statement holds for an initial value, generally \( n = 1 \) or \( n = 0 \). In the induction step, you assume the statement holds for an arbitrary positive integer \( k \) and then prove it holds for \( k + 1 \). The combination of these steps demonstrates that the statement is true for all natural numbers. It's like a domino effect; if one statement is true, the next one must be true, and so on.

Polynomial Division

Polynomial division is used to divide one polynomial by another, somewhat analogously to long division with numbers. Begin by dividing the leading term of the dividend by the leading term of the divisor. Then multiply the entire divisor by this quotient and subtract the result from the original polynomial. The remainder becomes the new dividend, and this process is repeated until the degree of the remainder is less than the degree of the divisor. The initial polynomial is then expressed as the divisor polynomial multiplied by the quotient polynomial, plus the remainder. This method often helps factor or simplify complex polynomials.

Proof by Induction

Proof by induction involves proving a base case and then proving that if a statement holds for an arbitrary case \( k \), it also holds for the next case \( k + 1 \). Let's apply this to show that \( a - b \) is a factor of \( a^n - b^n \):

• Base Case: For \( n = 1 \), the polynomial \( P(a, b) = a^1 - b^1 = a - b \). Clearly, \( a - b \) is a factor.
• Induction Hypothesis: Assume it's true for \( n = k \), i.e., \( a^k - b^k = (a - b)Q(a, b) \) for some polynomial \( Q(a, b) \).
• Inductive Step: Prove it for \( k + 1 \) using the given hint:

Using the hint, rewrite \( a^{k+1} - b^{k+1} = a(a^k - b^k) + b^k(a - b) \). Employing the induction hypothesis, substitute \( a(a^k - b^k) = a((a - b)Q(a, b)) \). Factoring out \( a - b \), we get \( a^{k+1} - b^{k+1} = (a - b)(aQ(a, b) + b^k) \), proving that \( a - b \) is indeed a factor for \( n = k + 1 \). Therefore, by induction, the statement holds for all \( n \).