# Componendo et Dividendo

This week, we learn about the method of Componendo et Dividendo.

How would you use Componendo et Dividendo to solve the following?

If $n$ is an integer, how many complex solutions are there to $\frac{ x^n+1} { x+1 } = \frac{ x^n -1 } { x-1}?$ Note by Calvin Lin
6 years, 11 months ago

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Assuming that $n \ge 2$, we manipulate the equation to obtain $\frac{x^n+1}{x^n-1} \;=\; \frac{x+1}{x-1}$ Thus, using Componendo et Dividendo with $k=1$ we deduce that $x^n \; =\; \frac{2x^n}{2} \; =\; \frac{(x^n+1)+(x^n-1)}{(x^n+1)-(x^n-1)} \; = \; \frac{(x+1)+(x-1)}{(x+1)-(x-1)} \; = \; \frac{2x}{2} \; = \; x$ and hence $x(x^{n-1}-1)=0$. We need to exclude both $x=1$ and $x=-1$ as solutions. Thus

• There are $n-1$ roots when $n$ is even, namely $0$ and the $n-2$ complex $(n-1)$st roots of unity ($1$ omitted),

• There are $n-2$ roots when $n$ is odd, namely $0$ and the $n-3$ complex $(n-1)$st roots of unity ($1$ and $-1$ omitted).

If $n=1$ then there are infinitely many roots.

If $n=0$ then there are no solutions.

If $n < 0$ then the same algebraic steps as above hold, but we have to exclude $0$ as a root, so the roots are the $|n-1|$th roots of unity, with $1$ and $-1$ excluded as necessary.

- 6 years, 11 months ago

You did a great job with the cases. This question was tricky, in part because you have to be actively aware of the different conditions that were necessary for the arguments to hold.

It also shows why after manipulating an equation, you must always check that the values hold in the original equation. This is a step that is often forgotten, with students claiming that "a degree $n$ polynomial must have $n$ roots."

With $n < 0$, most will also forget that they have to exclude 0 as a root. In this case, when $n\leq 3$ is an odd integer, then there are $|n| - 3$ roots. It is slightly interesting that for $n=-3$, we have no solutions, because all the possible solutions have to be rejected.

Staff - 6 years, 11 months ago

If $n=0$ then how can there be no solutions?

- 6 years, 11 months ago

also if n=0, x^n-1 = 0. (x^n + 1)/0 = math error kkk

- 6 years, 11 months ago

Yes, in the adjusted equation. However, when $n=0$, the first manipulation is not valid, and the arguments about $n=0$ are applied to the original equation. See Daniel C's comment.

- 6 years, 11 months ago

If $n=0$, we get $\dfrac{2}{x+1}=0$, which is not possible.

- 6 years, 11 months ago

Oh! yes, I forgot that there is no power of $n$ in the denominator. Funny that.

- 6 years, 11 months ago

$\frac{x^n+1}{x^n-1}=\frac{x+1}{x-1}$ Note that $x \neq 1, -1$.

Apply dividendo to get:

$\frac{2}{x^n - 1} = \frac{2}{x-1}$

$x(x^n-1)=0$

Suppose if $n$ is an even positive integer then we know that the polynomial[$x^n-1)$] has $n$ roots [not necessarily distinct]. Out of these $n$ roots, two are $1, -1$, so there are a total of $n-2+1=n-1$ solutions when $n$ is an even positive integer. When $n$ is odd, there $n-1$[except for $n=1$] solutions.

Note: It still needs to be shown that $x$ must be distinct for all those solutions.

Note that I have ignored distinct solutions for $x$.

If $n=0$ there are no solutions to $x$.

If $n$ is an odd positive integer, then there will be $n-1+1=n$ solutions to the equation, since $-1$ wont be a root in this case.

If $n$ is a non-positive integer, we let $n=-k$ for some positive integer $k$ and we get:

$x^{k+1} - 1 =0$

If $k$ is odd we have $k-1$[ $1,-1$ are not solutions] solutions, and if $k$ is even there are $k$[$k+1$ roots out of which $-1$ is not a solution] solutions.

- 6 years, 11 months ago

Indeed. This is an example of how manipulating the equation, leads to a different set of solutions. For example, both of you dealt with

$\frac{x^n + 1}{ x^n -1 } = \frac{x+1}{x-1}$

When $n$ is odd, we have $x = -1$ as a solution. However, this case has to be rejected in my version, and it is good that you mentioned "Note that $x \neq 1, -1$.

However, you should also have added the condition that $x^n -1 \neq 0$, in order to allow you to divide by the term. It is a quick check that if $x^n = 1$, then there are no solutions to the equation, since we must have

$\frac{ 2 }{x+1} = 0$

Staff - 6 years, 11 months ago

Typo:

$\text{1. Componendo:} \quad \quad \quad \quad \quad \frac{a+b}{b} = \frac{c-d}{d}.$

- 6 years, 11 months ago