# A smarter (not bashy) way to solve this roots of unity problem?

(Mandelbrot) Let $$\xi = \cos \frac{2\pi}{5} + i \sin \frac{2pi}{5}$$ be a complex fifth root of unity. Set $$a = 20\xi^2 + 13 \xi, b = 20\xi^4 + 13\xi^2, c = 20\xi^3 + 13\xi^4, \text{and } d = 20\xi + 13\xi^3$$. Find $$a^3 + b^3 + c^3 + d^3$$

Immediately what comes to mind is finding $$(a + b + c + d)^3$$ and subtracting whatever we don't need to get $$a^3 + b^3 + c^3 + d^3$$. However,

$$\begin{equation*} (a+b+c+d)^3 = a^3+3a^2b+3a^2c+3a^2d+3ab^2+6abc+6abd+3ac^2+6acd+3ad^2+b^3+3b^2c+3b^2d+3bc^2+6bcd+3bd^2 + c^3 + 3c^2d + 3cd^2 + d^3 \end{equation*}$$ There is simply no good way to calculate $$6abc + 6abd + 6acd + 6bcd$$ without expanding everything.

• $e^{ix}=\cos(x)+i\sin(x)$
– tp1
Oct 9, 2022 at 23:22
• Just asking: You've noticed $c = \bar{a}$ and $d = \bar{b}$? Oct 9, 2022 at 23:25
• seems like you can use $x^3+y^3 = (x+y)^3 - 3xy(x+y)$ for the conjugate pairs mentioned in the comment above. Oct 9, 2022 at 23:28
• Do you know about trace (in the context of field extensions)? Oct 9, 2022 at 23:29

You're not using the most important parts of the question: namely that $$\xi$$ is a fifth root of unity, and also that conveniently, all the coefficients are either $$20$$ and $$13$$.

Calculating $$a^3 + b^3 + c^3 + d^3$$ directly and collecting like coefficients, we have:

$$a^3 + b^3 + c^3 + d^3 = 20^3(\xi^6 + \xi^{12} + \xi^9 + \xi^3) + 3 \cdot 20^2 \cdot 13(\xi^5 + \xi^{10} + \xi^{10} + \xi^5)$$ $$+ 3 \cdot 20 \cdot 13^2 (\xi^4 + \xi^8 + \xi^{11} + \xi^{7}) + 13^3 (\xi^3 + \xi^6 + \xi^{12} + \xi^9)$$

and since $$\xi^5 = 1$$:

$$= 20^3 \left(\frac{1 - \xi^{15}}{1 - \xi^3} - 1 \right) + 3 \cdot 20^2 \cdot 13 \cdot4 +3 \cdot 20 \cdot 13^2 (\xi^4 + \xi^3 + \xi^1 + \xi^2) + 13^3 \left(\frac{1 - \xi^{15}}{1 - \xi^3} - 1 \right)$$

Can you continue from here?

• I know our answers don't agree but I'm poring over to see where I made a mistake. Oct 9, 2022 at 23:35
• Yep, I don't see anything wrong with $20^3 \cdot -1 + 3 \cdot 20^2 \cdot 13 \cdot 4 + 3 \cdot 20 \cdot 13^2 \cdot (-1) + 13^3 \cdot -1 = 42 \ 063$. Oct 9, 2022 at 23:41
• If these are directed towards me, I agree. I got the same result (after I corrected my own error). +1 Oct 9, 2022 at 23:42
• +1. Note: Since $\xi^5=1$, three of the $\xi$-sums immediately reduce to $\xi+\xi^2+\xi^3+\xi^4$ (which is $-1$). No need for the geometric sum formula with the first and last terms.
– Blue
Oct 9, 2022 at 23:57
• Yes, there are different ways of thinking about it. Since the roots of unity are vertices of a regular pentagon, their sum must be $0$; but also the geometric sum method is also helpful for thinking about relationships between complex numbers as multiplications. Oct 10, 2022 at 0:00

Let $$\omega_1, \ldots, \omega_p$$ be the $$p$$th roots of unity (including $$1$$), where $$p$$ is prime. Note that: $$\omega_1^k + \ldots + \omega_p^k = \begin{cases} p & \text{if } p \mid k \\ 0 & \text{otherwise.} \end{cases}$$ Why? The group of $$p$$th roots of unity are isomorphic to $$\Bbb{Z} / p\Bbb{Z}$$, and the map $$x \mapsto x^k$$ is equivalent to the map $$x \mapsto kx$$ on $$\Bbb{Z} / 5 \Bbb{Z}$$. It's well know that this homomorphism is trivial when $$p \mid k$$, and injective (and thus surjective) otherwise. So, when $$p \not\mid k$$, we are summing the $$p$$th roots of unity, which is $$0$$.

So, if we have a polynomial $$q$$, we can evaluate $$q(\omega_1) + \ldots + q(\omega_p)$$ relatively easily. We get: $$q(\omega_1) + \ldots + q(\omega_p) = p(q_0 + q_p + q_{2p} + \ldots),$$ where $$q_i$$ is the coefficient of $$x^i$$, and the sum on the right hand side is necessarily finite, due to the finite degree of $$q$$.

In your case, where $$p = 5$$, we have $$a^3, b^3, c^3, d^3, (20 + 13)^3$$ are (not necessarily respectively) $$q(\omega_1), \ldots, q(\omega_5)$$, where $$q(x) = (20x^2 + 13x)^3.$$ We only care about the coefficients of $$x^5$$ and $$x^0$$, the latter of which is clearly $$0$$, and the former is $$3 \cdot 20^2 \cdot 13$$. So, $$a^3 + b^3 + c^3 + d^3 + 33^3 = 15 \cdot 20^2 \cdot 13,$$ hence the answer is $$15 \cdot 20^2 \cdot 13 - 33^3$$

They simplified exponents when defining $$\xi,$$ but you could also write the equations \begin{align*}a &= \xi(20\xi+13),\\b &= \xi^2(20\xi^2+13),\\c &= \xi^3(20\xi^3+13),\\d &= \xi^4(20\xi^4+13)\end{align*} (actually, they swapped $$c$$ and $$d$$, but that doesn't change the problem). Consider the polynomial $$f(x) = [x(20x+13)]^3 = 20^3x^6+3\cdot20^2\cdot13x^5+3\cdot20\cdot13^2x^4+13^3x^3.$$ A roots of unity filter* using fifth roots of unity would give the sum of the coefficients of $$x^0, x^5, x^{10},$$ etc., which is just $$3\cdot20^2\cdot 13 = 15600.$$ But $$\xi$$ is a fifth root of unity! So, $$15600 = \frac{f(1)+f(\xi)+f(\xi^2)+f(\xi^3)+f(\xi^4)}{5} = \frac{33^3+a^3+b^3+c^3+d^3}{5}.$$ Rearranging gives $$a^3+b^3+c^3+d^3=78000-33^3=\boxed{42063}.$$

*Say we have some polynomial $$P(x) = a_0x^0+a_1x^1+a_2x^2+\cdots+a_nx^n$$ Plugging in powers of the $$k$$th root of unity $$\omega$$ gives $$\sum_{j=0}^{k-1}P(\omega^j) = \sum_{i=0}^n a_i\sum_{j=0}^{k-1}(\omega^{j})^i.$$ Inspecting the inner summand, we find $$\sum_{j=0}^{k-1}(\omega^i)^j = k$$ if $$\omega^i = 1$$ (i.e. $$i$$ is a multiple of $$k$$), otherwise $$(\omega^i-1)\sum_{j=0}(\omega^i)^j = (\omega^{i})^k-1 = 0,$$ so it evaluates to zero.

Let $$\varphi:\mathbb{Q}[\xi]\to\mathbb{Q}[\xi]$$ be the field homomorphism which is the identity on $$\mathbb{Q}$$ and $$\varphi(\xi)=\xi^2$$.

Since $$\xi+\xi^2+\xi^3+\xi^4=-1$$, we have \begin{align} a^3&=\left(20\xi^2+13\xi\right)^3&&=8000\xi+15600+10140\xi^4+2197\xi^3\\ b^3&=\varphi\!\left(a^3\right)&&=8000\xi^2+15600+10140\xi^3+2197\xi\\ c^3&=\varphi\circ\varphi\!\left(a^3\right)&&=8000\xi^4+15600+10140\xi+2197\xi^2\\ d^3&=\varphi\circ\varphi\circ\varphi\!\left(a^3\right)&&=8000\xi^3+15600+10140\xi^2+2197\xi^4\\\hline &\!\!\!\!\!a^3+b^3+c^3+d^3&&=-8000+62400-10140-2197\\ &&&=42063 \end{align}