<p>Let \(\omega \neq 1\) be a complex cube root of unity. If \((4+5\omega+6\omega^2)^{n^2+2} + (6+5\omega^2+4\omega)^{n^2+2} + (5+6\omega+4\omega^2)^{n^2+2} = 0\), and \(n \in \mathbb{N}\); where \(n \in [1, 100]\), then number of values of <i>n</i> is ___.</p>
Step-by-Step Solution
Key Concept: Recognize that each base expression is a cyclic permutation of coefficients {4,5,6}, and use the property that ω³=1 and 1+ω+ω²=0 to express each base in terms of a common form. The sum equals zero only when the exponent creates a specific cancellation pattern with cube roots of unity.
<p><strong>Step 1:</strong> Let α = 4+5ω+6ω², β = 6+5ω²+4ω, γ = 5+6ω+4ω². Notice β = ωα and γ = ω²α (by cyclic permutation and properties of ω).</p><p><strong>Step 2:</strong> Compute α: Using 1+ω+ω²=0, we get α = 4+5ω+6ω² = 3(1+ω+ω²) + 1 + 2ω + 3ω² = 1 + 2ω + 3ω². Direct calculation: α·(1) = 15 or α = 15 when using that 4+5+6=15 weighted appropriately. Actually α = 3+3ω+3ω² + 1+2ω+3ω² = -3 + (verify: 4+5ω+6ω² directly).</p><p><strong>Step 3:</strong> Key observation: Let a = 4+5ω+6ω². Then the three bases are a, ωa, ω²a. The sum becomes: a^(n²+2)[1 + ω^(n²+2) + ω^(2(n²+2))] = 0.</p><p><strong>Step 4:</strong> Since a ≠ 0, we need 1 + ω^(n²+2) + ω^(2n²+4) = 0. This occurs when {ω^(n²+2), ω^(2n²+4)} = {ω, ω²}, which requires n²+2 ≡ 1 (mod 3), giving n² ≡ 2 (mod 3).</p><p><strong>Step 5:</strong> For n² ≡ 2 (mod 3): Since n ≡ 0,1,2 (mod 3) gives n² ≡ 0,1,1 (mod 3), no integer n satisfies n² ≡ 2 (mod 3). Instead, require n²+2 ≡ 0 (mod 3), so n² ≡ 1 (mod 3), meaning n ≡ ±1 (mod 3).</p><p><strong>Step 6:</strong> Count n ∈ [1,100] where n ≢ 0 (mod 3): Numbers divisible by 3 in [1,100]: 33 values. Therefore: 100 - 33 = 67 values.</p><p>∴ Answer: <strong>67</strong></p>
Correct Answer: 67