Find the sum of the roots, real and non-real, of the equation $x^{2001}+\left(\frac 12-x\right)^{2001}=0$, given that there are no multiple roots.

While trying to solve the above problem (AIME 2001, Problem 3) which was asked here on MSE, I came across three solutions on AoPS. ( The MSE solution uses Vieta's formula which I am clear about )

The first solution (on AoPS) involves the use of Vieta's formula's and is quite clear.

The third solution states the following :

Note that if $r$ is a root, then $\frac{1}{2}-r$ is a root and they sum up to $\frac{1}{2}.$

We make the substitution $y=x-\frac{1}{4}$ so $(\frac{1}{4}+y)^{2001}+(\frac{1}{4}-y)^{2001}=0.$

Expanding gives $2\cdot\frac{1}{4}\cdot\binom{2001}{1}y^{2000}-0y^{1999}+\cdots$ so by Vieta, the sum of the roots of $y$ is 0.

Since $x$ has a degree of 2000, then $x$ has 2000 roots so the sum of the roots is $2000(\sum_{n=1}^{2000} y+\frac{1}{4})=2000(0+\frac{1}{4})=\boxed{500}.$

I do not understand two things in the above solution :

  1. "Note that if $r$ is a root, then $\frac{1}{2}-r$ is a root and they sum up to $\frac{1}{2}.$"

a) Here what is being referred to as "they"? (Shouldn't $\frac{1}{2}-r$ be a factor and not a root). Answered by the third comment

b)Why is the sum 1/2? Also answered by the third comment

c)Why is $\frac{1}{2}-r$ a root? Answered by the sixth comment

  1. How is the final expression arrived upon to find the sum of all 2000 roots? (Answered by @YvesDaoust)

The second solution is more mystifying (possibly because it is similar to the one above):

We find that the given equation has a $2000^{\text{th}}$ degree polynomial. Note that there are no multiple roots. Thus, if $\frac{1}{2} - x$ is a root, $x$ is also a root. Thus, we pair up $1000$ pairs of roots that sum to $\frac{1}{2}$ to get a sum of $\boxed{500}$.

  1. Again, why is $\frac{1}{2} - x$ a root. By "$x$ is also a root" does it mean $x$ representing the set of all roots? Answered by the sixth comment

  2. Why does the pairing up occur? Why is the sum of each pair 1/2? (Answered by @YvesDaoust)

I wonder if it could be solved as follows :

Let the roots be $P_1,P_2,...P_{2000}$. The polynomial can be expressed as a product of factors as follows : $(2001/2)($x$-P_1)($x$-P_2)....($x$-P_{2000}) = 0$. The above expression is the same as $x^{2001}+\left(\frac 12-x\right)^{2001}=0$.

Thus, $x^{2001}+\left(\frac 12-x\right)^{2001}$ = $(2001/2)($x$-P_1)($x$-P_2)....($x$-P_{2000})$

Here the coefficient of $x^{1999}$ on the $RHS$ should represent $\sum\limits_{i=1}^{2000}P_i$$\times$$(-1/2)$

On the $LHS$ the corresponding term would be the term with $x^{1999}$ and thus the coefficient of this term on the $LHS$ should also be the required sum.

On the LHS the coefficient of the $x^{1999}$ term is -${2001}\choose{2}$*$(1/2)^2$ which represent the sum of the roots.

[Picked up this approach here, but I don't see how this would work ] (https://youtu.be/S6FRtmDUl-s?t=2806)

In this solution I find some errors(?) :

  1. Are there any inconsistencies in the reasoning?Wouldn't the sum of roots differ from the binomial coefficient since the expression involves unique values of $P_i$ (no multiple roots).
  2. The answers do not match, which seems to suggest so.
  3. Is there a way of arriving at the answer without using Vieta's formula and by expressing the polynomial as a product of factors and then using binomial coefficients as attempted above?
  • 4
    $\begingroup$ That's seven questions! $\endgroup$ Commented Oct 13, 2018 at 5:00
  • $\begingroup$ The first solution you give (which you call the third solution) is essentially the same as the second solution you give (which mercifully you call the second solution). $\endgroup$ Commented Oct 13, 2018 at 5:01
  • 2
    $\begingroup$ In your question 1, are you asking why $r+\left(\frac12-r\right)=\frac12$? $\endgroup$ Commented Oct 13, 2018 at 5:03
  • 2
    $\begingroup$ $\dfrac12-r$ is a root whenever $r$ is. This is because $$P(\dfrac12-r)=(\dfrac12-r)^{2001}-(\dfrac12-(\dfrac12-r))^{2001}= (\dfrac12-r)^{2001}-r^{2001}=-P(r).$$ So if $P(r)=0$ then $P(\dfrac12-r)=0$ also. $\endgroup$ Commented Oct 13, 2018 at 7:26
  • 1
    $\begingroup$ You're right. I made a sign error. Anyway, the argument, as per your modification, survives with the correct sign as well. $\endgroup$ Commented Oct 20, 2018 at 20:28

2 Answers 2


First solution:

By symmetry of $x^n+(\frac12-x)^n$, if $x$ is a root, so is $\frac12-x$. Now if you take the roots in pairs ($1000$ pairs), the sum of the individual pairs is $x+\frac12-x=\frac12$. Hence in total $1000\cdot\dfrac12$.

Second solution:

There is no difference with the first.

Extra solution:

By Vieta, the sum of the roots is the negative ratio of the two coefficients of the highest degree. Then using the Binomial theorem,

  • degree $2001$: $1-1=0$,

  • degree $2000$: $\dfrac12\dfrac{2001}2\left(-\dfrac12\right)$,

  • degree $1999$: $\dfrac{2001}2\left(\dfrac{2001-1}2\right)\left(-\dfrac12\right)^2$.

The requested ratio is indeed $500$.

  • $\begingroup$ Yes, that answers question 2 and 4! I think the extra solution is the same as the first solution on $[AoPS](artofproblemsolving.com/wiki/… whose working I did not add. @YvesDaoust $\endgroup$
    – idunno
    Commented Oct 13, 2018 at 9:47

In answer to questions 5,6,7 (the attempted solution) :

  1. It was explained to me elsewhere that the leading coefficient on the $RHS$ should not be $(1/2)$ but in fact be $(2001/2)$ since the leading coefficient for the polynomial of order 2000 would be ${2001}\choose{2000}$$\times$$(1/2)$.

  2. Thus on solving for the sum of the roots the answer would be $500$.

  3. It was also pointed out to me that this would essentially be the same as using Vieta's.

(Apologies for cross posting, I just saw a meta question regarding this and realized my error. I intend to follow the guidelines henceforth).


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