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Algebra Difficulty 4.7 AIME Prove it Croatia

Let aa be a complex number such that
a5+a+1=0.a^5 + a + 1 = 0.
What values can the expression a2(a1)a^2 (a - 1) take on?

Solution

We have
a5+a+1=(a5a4)+(a4a3)+(a3a2)+a2+a+1=(a2+a+1)(a3a2+1). \begin{aligned} a^5 + a + 1 &= (a^5 - a^4) + (a^4 - a^3) + (a^3 - a^2) + a^2 + a + 1 \\ &= (a^2 + a + 1)(a^3 - a^2 + 1). \end{aligned}
If a2+a+1=0a^2 + a + 1 = 0, then a2(a1)=(a+1)(a1)=1a2=a+2a^2(a - 1) = -(a + 1)(a - 1) = 1 - a^2 = a + 2.
Since a=1±i32a = \frac{-1 \pm i\sqrt{3}}{2}, it follows that a2(a1)=32±32ia^2(a - 1) = \frac{3}{2} \pm \frac{\sqrt{3}}{2} i.
If a3a2+1=0a^3 - a^2 + 1 = 0, then a2(a1)=a3a2=1a^2(a - 1) = a^3 - a^2 = -1.
Therefore, the observed expression can take on values 1,32,32i-1, \frac{3}{2}, \frac{\sqrt{3}}{2} i and 3232i\frac{3}{2} - \frac{\sqrt{3}}{2} i.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.