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How many values of z (real or complex) satisfy both 1 + z + z² + z³ +... + z¹⁷ = 0 and 1 + z + z² +... + z¹³ = 0 simultaneously?

1. 1
2. 2
3. 3
4. 4

Correct answer: 1

Solution

The first sum is zero iff z¹⁸ = 1, z != 1; the second iff z¹⁴ = 1, z != 1. Common roots satisfy z^gcd(18,14) = z² = 1, i.e. z = 1 or z = -1; excluding z = 1 leaves only z = -1.

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