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2010 Paper 3 Q3
D: 1700.0 B: 1545.2
complex numbers roots of unity cyclotomic polynomials primitive roots polynomials proof by contradiction De Moivre

For any given positive integer \(n\), a number \(a\) (which may be complex) is said to be a primitive \(n\)th root of unity if \(a^n=1\) and there is no integer \(m\) such that \(0 < m < n\) and \(a^m = 1\). Write down the two primitive 4th roots of unity. Let \({\rm C}_n(x)\) be the polynomial such that the roots of the equation \({\rm C}_n(x)=0\) are the primitive \(n\)th roots of unity, the coefficient of the highest power of \(x\) is one and the equation has no repeated roots. Show that \({\rm C}_4(x) = x^2+1\,\).

  1. Find \({\rm C}_1(x)\), \({\rm C}_2(x)\), \({\rm C}_3(x)\), \({\rm C}_5(x)\) and \({\rm C}_6(x)\), giving your answers as unfactorised polynomials.
  2. Find the value of \(n\) for which \({\rm C}_n(x) = x^4 + 1\).
  3. Given that \(p\) is prime, find an expression for \({\rm C}_p(x)\), giving your answer as an unfactorised polynomial.
  4. Prove that there are no positive integers \(q\), \(r\) and \(s\) such that \({\rm C}_q(x) \equiv {\rm C}_r(x) {\rm C}_s(x)\,\).

Solution: The primitive 4th roots of unity are \(i\) and \(-i\). (Since the other two roots of \(x^4-1\) are also roots of \(x^2-1\) \({\rm C}_4(x) = (x-i)(x+i) = x^2+1\) as required.

  1. \(\,\) \begin{align*} && {\rm C}_1 (x) &= x-1 \\ && {\rm C}_2 (x) &= x+1 \\ && {\rm C}_3 (x) &= x^2+x+1 \\ && {\rm C}_5 (x) &= x^4+x^3+x^2+x+1 \\ && {\rm C}_6 (x) &= x^2-x+1 \\ \end{align*}
  2. Since \((x^4+1)(x^4-1) = x^8-1\) we must have \(n \mid 8\). But \(n \neq 1,2,4\) so \(n = 8\).
  3. \({\rm C}_p(x) = x^{p-1} +x^{p-2}+\cdots+x+1\)
  4. Suppose \({\rm C_q}(x) \equiv {\rm C}_r(x){\rm C}_s(x)\), then if \(\omega\) is a primitive \(q\)th root of unity we must \({\rm C}_q(\omega) = 0\), but that means that one of \({\rm C}_r(\omega)\), \({\rm C}_s(\omega)\) is \(0\). But that's only possible if \(r\) or \(s\) \(=q\). If this were the case, then what would the other value be? There are no possible values, hence it's not possible.

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