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For a natural number n, which of the following always divides the expression x^(n+1) + (x + 1)^(2n−1)?

1. x
2. x + 1
3. x² + x + 1
4. x² − x + 1

Correct answer: x² + x + 1

Solution

Let w be a root of x^2+x+1=0, so w^3=1 and x+1 = -w^2. Then x^(n+1)+(x+1)^(2n-1) = w^(n+1) - w^(4n-2); since 4n-2 ≡ n+1 (mod 3), the two terms cancel, giving 0. Hence x^2+x+1 always divides it (x does not, since at x=0 the value is 1).

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