How do I solve a problem with term $a^{n} + b^{n}$?

Question

Given two non-zero numbers $x$ and $y$ such that $x^{2} + xy + y^{2} = 0$.

Find the value of $$\left(\frac{x}{x + y}\right)^{2013} + \left(\frac{y}{x + y}\right)^{2013}$$.

I found out that $(x + y)^2 = xy$ and I'm stuck at $\frac{x^{2013} + y^{2013}}{(x + y)^{2013}}$

Does anyone know how to solve this?

Answer 1

Since $y=x\exp\frac{\pm2\pi i}{3}$,$$\frac{x^n+y^n}{(x+y)^n}=\frac{1+\exp\frac{\pm2\pi i n}{3}}{(1+\exp\frac{\pm2\pi i}{3})^n}=\frac{2\exp\frac{\pm\pi i n}{3}\cos\frac{\pi n}{3}}{(2\exp\frac{\pm\pi i}{3}\cos\frac{\pi}{3})^n}=2\cos\frac{\pi n}{3}.$$In the case $n=2013$, this simplifies to $-2$ because $n/3$ is odd.

Answer 2

Hint: $x^2+xy+y^2=0$ leads to $(\frac{x}{y})^2+\frac{x}{y}+1=0$, which is a quadratic equation about $\frac{x}{y}$

Answer 3

Let $a_n = \left(\dfrac{x}{x + y}\right)^{n} + \left(\dfrac{y}{x + y}\right)^{n}$.

Since $$ \dfrac{x}{x + y}+\dfrac{y}{x + y}=1, \quad \dfrac{x}{x + y}\cdot\dfrac{y}{x + y}=1 $$ they are the roots of $t^2=t-1$ and so we get $$ a_{n+2} = a_{n+1}-a_n, \quad a_0=2, \quad a_1=1 $$ This sequence is periodic of period $6$: $$ 2,1,-1,-2,-1,1, 2,1,-1,-2,-1,1, \dots $$ Thus, $a_{2013} = a_{2013 \bmod 6} = a_3 = -2$.

Answer 4

Hint:

Clearly $xy\ne0$

$$\left(\dfrac xy\right)^2+\dfrac xy+1=0$$

$\implies \dfrac xy=w$ where $w$ is a complex cube root of unity $\implies w^3=1$

$$\dfrac x{x+y}=\dfrac{\dfrac xy}{1+\dfrac xy}=\dfrac w{1+w}=\dfrac w{-w^2}=-w^2$$

$$\dfrac y{x+y}=\dfrac1{1+\dfrac xy}=\dfrac1{1+w}=\dfrac1{-w^2}=-w$$

Answer 5

Another variant: set $t=\dfrac y x$. Then $t$ satisfies the equation $1+t+t^2=0$, i.e. $t$ is one of the complex cubic roots of unity $j, j^2$, and for any $n$, we have

$$\biggl(\frac{x}{x+y}\biggr)^{\!n}+\biggl(\frac{y}{x+y}\biggr)^{\!n}=\frac{x^n(1+t^n)}{x^n(1+t)^n}=\frac{1+t^n}{(1+t)^n}$$ There remains to use that $\;1+t=-t^2=-\bar t$ and $t^n=t^{n\bmod 3}$.

Answer 6

$x^3-y^3=(x-y)(x^2+xy+y^2)=0$ $\implies$ $x^3=y^3$

$\left(\dfrac x{x+y}\right)^{2013}=\left(\dfrac{x^2}{x^2+xy}\right)^{2013}=\left(\dfrac{x^2}{-y^2}\right)^{2013}=-\left(\dfrac{x^3}{y^3}\right)^{1342}=-1$

Similarly, $\left(\dfrac y{x+y}\right)^{2013}=-1$.

The sum is $-2$.

Answer 7

We have $${y}^{2}=-{x}^{2}-xy\\ {y}^{3}=-{x}^{2}y-xy^2 $$ Then after substition and elimination $y, y^2,y^3$ we get
$$ (x+y)^3={x}^{3}+3\,{x}^{2}y+3\,x{y}^{2}+{y}^{3}=-x^3. $$ Thus $$ \left(\dfrac x{x+y}\right)^{2013}=\left(\left(\dfrac x{x+y}\right)^{3}\right)^{671}=\left(\dfrac{ x^3}{(x+y)^3}\right)^{671}=\left(\dfrac{ x^3}{-x^3}\right)^{671}=-1. $$ Similarly $$\left(\dfrac y{x+y}\right)^{2013}=-1.$$