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Chapter 6: Problem 469

If \(\mathrm{x}^{2 \mathrm{n}-1}+\mathrm{y}^{2 \mathrm{n}-1}\) is divisible by \(\mathrm{x}+\mathrm{y}\), then \(\mathrm{n}\) is (a) Positive integer (b) Only for an even positive integer (c) An odd positive integer (d) \(\forall \mathrm{n}, \mathrm{n} \geq 2\)

Short Answer

Expert verified
The correct answer is (a) Positive integer.

Step by step solution

01

Write the given expression

We are given that the expression \(x^{2n-1} + y^{2n-1}\) is divisible by \(x + y\).
02

Factorize using difference of squares formula

Recall the difference of squares formula: \(a^2 - b^2 = (a + b)(a - b)\). Although our expression is a sum, not a difference, we can apply a similar approach. For a start, let's rewrite our expression as follows: \[x^{2n-1} + y^{2n-1} = x^{(2n-2) + 1} + y^{(2n-2) + 1}\]
03

Apply Advanced Factorization Technique

We will now apply an algebraic identity known as Sophie Germain Identity, which is given by: \[(a^4 + b^4 + 4c^4 - 2a^2b^2 - 2a^2c^2 - 2b^2c^2) = (a^2 + b^2 + c^2 - abc)^2\] In our expression, we can rewrite this identity as follows with a=x, b=y, and c=1: \[x^{4n-4}+y^{4n-4}+4-2x^{2n-2}y^{2n-2}-2x^{2n-2}-2y^{2n-2}=(x^{2n-2}+y^{2n-2}+1-xy^{2n-2})^2\] Now, let's check if our expression fits the Sophie Germain Identity.
04

Check divisibility with Sophie Germain Identity

We need to show that \(x^{2n-1} + y^{2n-1}\) is divisible by \(x + y\). This would be true if it matches the left-hand side of the Sophie Germain Identity. Multiplying \(x+y\) with our expression for the left-hand side of the identity, we get: \[(x+y)(x^{4n-4}+y^{4n-4}+4-2x^{2n-2}y^{2n-2}-2x^{2n-2}-2y^{2n-2})\] Since we notice that our expression has the form (a+b)(something), it only holds when n is a positive integer as in choice (a). Thus, the correct answer is: (a) Positive integer

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Most popular questions from this chapter

The greatest positive integer, which divides \((\mathrm{n}+2)(\mathrm{n}+3)\) \((n+4)(n+5)(n+6)\) for all \(\forall n \in N\), is (a) 120 (b) 4 (c) 240 (d) 24

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