. . . . . . . . . . . . . . . . . . . . . . . . . . Olympiad
Training for Individual Study
Symmetric Polynomials
ft. Vieta’s polynomials
Evan Chen《陳誼廷》
6 April 2023
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BAW-SYMPOLY
OTIS, © Evan Chen, internal use only. Artwork contributed by Rohan Dhillon.
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Evan Chen《陳誼廷》 (OTIS, updated 2023-04-06)
Symmetric Polynomials
§1 Lecture notes
A lot of the intuition you’ll get from this lecture is hard to state formally, but there are a
few results you should be aware of off the bat.
• Vieta: the coefficients of a polynomial are symmetric sums.
• Factoring theorem: again two things here.
(a) If P (x) is a polynomial with root r, then x − r divides P . Easy.
(b) If P (x, y, . . . ) is a polynomial vanishing whenever x = y, then x − y divides P .
§1.1 Fundamental theorem of elementary symmetric polynomials
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More involved is the fundamental theorem of symmetric polynomials: Let
P (x1 , . . . , xn ) be a polynomial which is symmetric. Consider the so-called Vieta basis:
X
σ1 =
xi
i
σ2 =
X
xi xj
i<j
σ3 =
..
.
X
xi xj xk
i<j<k
σn = x1 . . . xn
(These are just the formulas you get from Vieta formulas, without the pesky signs.)
Theorem 1.1 (Fundamental theorem of elementary symmetric polynomials)
Any symmetric polynomial P (x1 , . . . , xn ) can be expressed uniquely as a polynomial
in σ1 , . . . , σn . That is,
P (x1 , . . . , xn ) = Q(σ1 , . . . , σn )
for some unique polynomial Q.
Moreover, if P has integer/rational/real coefficients then so does Q.
As an example of the previous theorem, this means that it should possible to express
a3 + b3 + c3 + d3 in terms of the symmetric sums σ1 = a + b + c + d, σ2 = ab + bc + cd +
da + ac + bd, σ3 = abc + bcd + cda + dab, σ4 = abcd. And indeed it is:
a3 + b3 + c3 + d3 = (a + b + c + d)3 − 3(a + b + c + d)(ab + bc + cd + da + ac + bd)
+ 3(abc + bcd + cda + dab)
= σ13 − 3σ1 σ2 + 3σ3 .
This means that the Vieta basis is sort of the natural one in which one can “force” a
canonical representation of a symmetric polynomial.
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Evan Chen《陳誼廷》 (OTIS, updated 2023-04-06)
Symmetric Polynomials
§1.2 Factoring multivariate polynomials
When you have a single variable polynomial P (t) with a root at t = r, then you know
that t − r is a factor of P , and depending on context you can sometimes use this to
extract factorizations of P .
A similar statement is true for multivariate polynomials given a linear relation like
P = 0 when a = b or P = 0 when a + b + c = 0. (Although we won’t use it, a slightly
more technical result is true for higher-degree relations; https://math.stackexchange.
com/a/463935/229197 if interested.) But when symmetric, you enjoy a lot of additional
symmetry which you can often exploit.
For example, consider the polynomial
P (a, b, c) = a3 + b3 + c3 − 3abc.
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Notice that P has the property that it vanishes whenever a + b + c = 0. This means that
P should factor as the product of a + b + c and a degree-2 symmetric polynomial. In
other words, we can write
P (a, b, c) = a3 + b3 + c3 − 3abc = (a + b + c) λ1 (a2 + b2 + c2 ) + λ2 (ab + bc + ca)
Now you can deduce the constants λ1 and λ2 by examining just a few particular coefficients.
For example, equating the a3 coefficient terms on each side gives
a3 = a · (λ1 a2 ) =⇒ λ1 = 1
while equating the abc term gives
−3abc = λ2 (a · bc + b · ca + c · ab) =⇒ λ2 = −1.
§1.3 Newton sums, classical form
In the case where we want to express r1d + · · · + rnd in terms of the Vieta coefficients, the
Newton Sums lets us do this explicitly (or at least recursively). This doesn’t come up
too often but is nice to know.
Theorem 1.2 (Newton sums)
Consider the polynomial
P (x) = an xn + an−1 xn−1 + · · · + a0
with complex roots r1 , r2 , . . . , rn ,
For each d ≥ 0, define pd = r1d + r2d + · · · + rnd . Then
0 = an p1 + an−1
= an p2 + an−1 p1 + an−2 · 2
= an p3 + an−1 p2 + an−2 p1 + an−3 · 3
= an p4 + an−1 p3 + an−2 p2 + an−3 p1 + an−4 · 4
..
.
where we by convention let ai = 0 if i < 0.
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Evan Chen《陳誼廷》 (OTIS, updated 2023-04-06)
Symmetric Polynomials
To be concrete, consider the polynomial x3 + 7x2 + 14x + 8, say with roots u, v, w.
If we want to compute u3 + v 3 + w3 , then the Newton sums identities let us do this
recursively by
0 = 1 · p1 + 7 · 1 =⇒ p1 = −7
0 = 1 · p2 + 7 · p1 + 14 · 2 =⇒ p2 = 21
0 = 1 · p3 + 7 · p2 + 14 · p1 + 8 · 3 =⇒ p3 = −73.
§1.4 Newton sums, obscure forms (just for curiosity)
If you like determinants, you can use the following formula instead:
Theorem 1.3 (Newton’s identities in Cauchy form)
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Let σi be as defined last section.

σ1

2σ2


3σ3


4σ4
pd = det 

..

.

(d − 1)σd−1
dσd
For each d ≥ 0, we have
1
σ1
σ2
σ3
..
.
0
1
σ1
σ2
..
.
0
0
1
σ1
..
.
0
0
0
1
..
.
...
...
...
...
..
.
0
0
0
0
..
.
σd−2 σd−3 σd−4 σd−5 . . . σ1
σd−1 σd−2 σd−3 σd−4 . . . σ2

0
0

0

0

.. 
.

1
σ1
Note that this formula uses σi (power sums) whereas the last one uses ai (coefficients);
for monic polynomials, these are related by σi = (−1)i an−i . So to redo the previous
example we would write


−7
1
0
p3 = det  2 · 14 −7 1  = −73.
3 · (−8) 14 −7
And if you like sums, or you just want the explicit polynomial:
Theorem 1.4 (Newton’s identities as an explicit sum)
Let σi be as defined last section. For each d > 0, we have
X
(−1)s
s
d
pd = (−1) · d ·
σ r1 σ r2 . . . σdrd
s
r1 , . . . , rd 1 2
r1 +2r2 +···+drd =d
ri ≥0
where s = r1 + r2 + · · · + rd for brevity.
If you don’t like multinomial coefficients, you can also write out the factorials:
pd = (−1)d · d ·
X
r1 +2r2 +···+drd =d
ri ≥0
(−1)s (s − 1)! r1 r2
σ σ . . . σdrd .
r1 !r2 ! . . . rd ! 1 2
You do not have to memorize either of these obscure forms as they are not used
often. (Heck, I didn’t even know either existed until after I graduated college.) They’re
included here as a reference, not as mandatory reading.
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Evan Chen《陳誼廷》 (OTIS, updated 2023-04-06)
Symmetric Polynomials
§1.5 Examples
Factoring theorem example:
Example 1.5
Show that a4 + b4 ≥ a3 b + ab3 for a, b > 0.
For the purposes of this example, assume you don’t know AM-GM or Muirhead, etc.
The goal is to show how to solve the problem with “bare hands”. Walkthrough. We
will prove a4 + b4 − a3 b − ab3 ≥ 0.
(a) Noting that equality holds when a = b, what factor must divide the left-hand side?
(b) Imagine fixing b, and treating the left-hand side as a polynomial P (a). It has a
root at a = b. If you also know that P ≥ 0 everywhere, what kind of root must
that root be?
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(c) Use this to factor the left-hand side completely. (There should be three factors.)
The condition a, b > 0 isn’t actually used here but makes things simpler to think about.
Example 1.6 (AIME 2010)
Compute the maximum possible value of a3 + b3 + c3 given
a3 = abc + 2
b3 = abc + 6
c3 = abc + 20.
Walkthrough.
(a) Express a3 + b3 + c3 in terms of abc. Thus it suffices to compute abc.
(b) Find a way to get a quadratic equation in abc.
(c) Solve for abc and use it to get the final answer.
Example 1.7 (OMO Fall 2015)
Let a, b, c be the distinct roots of the polynomial
P (x) = x3 − 10x2 + x − 2015.
The cubic polynomial Q(x) is monic and has distinct roots bc − a2 , ca − b2 , ab − c2 .
What is the sum of the coefficients of Q?
Walkthrough. This can be done with brute force, by actually finding Q, but there is a
trick to it.
(a) Show the answer is given by Q(1) = (1 − bc + a2 )(1 − ca + b2 )(1 − ab + c2 ).
(b) Show that ab + bc + ca = 1.
(c) Prove that Q(1) = 2015000.
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Evan Chen《陳誼廷》 (OTIS, updated 2023-04-06)
Symmetric Polynomials
Example 1.8 (USAMO 1975/3)
If P (x) denotes a polynomial of degree n such that P (k) =
determine P (n + 1).
75AMO3
k
k+1
for k = 0, 1, 2, . . . , n,
Walkthrough. The main idea is to define Q(x) = (x + 1)P (x) − x.
(a) Compute deg Q (in terms of n).
(b) Determine the roots of Q.
(c) Use (a) and (b) to establish the factorization, of Q up to a constant factor.
(d) Show that Q(−1) = 1 and use this to conclude that the leading coefficient of Q is
equal to
(−1)n+1
c=
.
(n + 1)!
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(e) Compute Q(n + 1).
(f) Prove that P (n + 1) =
n+1+(−1)n+1
.
n+2
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Evan Chen《陳誼廷》 (OTIS, updated 2023-04-06)
Symmetric Polynomials
§2 Practice Problems
Instructions: Solve [30♣]. If you have time, solve [36♣]. Problems with red weights are mandatory.
The Law speaks: you are cast out. You are un-dwarf. I AM
A WITNESS!
Angarthing in The Hammer of Thursagan,
from The Battle for Wesnoth
73AMO4
[2♣] Problem 1 (USAMO 1973/4). Determine all triples (x, y, z) of complex numbers
satisfying
x + y + z = 3,
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2
x + y 2 + z 2 = 3,
x3 + y 3 + z 3 = 3.
96CAN1
16HMNTGUTS27
NIMO85
[2♣] Problem 2 (Canada 1996, added by Haozhe Yang). If α, β, and γ are the roots of
1+β
1+γ
x3 − x − 1 = 0, compute 1+α
1−α + 1−β + 1−γ .
[2♣] Problem 3 (HMMT November 2016 Guts, added by Rohan Bodke). Let r1 , r2 , r3 ,
r4 be the roots of the polynomial x4 − 4x3 + 8x2 − 7x + 3. Calculate
r12
r22
r32
r42
+
+
+
.
r22 + r32 + r42 r12 + r32 + r42 r12 + r22 + r42 r12 + r22 + r32
[2♣] Problem 4 (NIMO #8). Let x, y, z be complex numbers satisfying
x2 + 5y = 10x
y 2 + 5z = 10y
z 2 + 5x = 10z.
84AMO1
JASONMAO
20AIMEII11
16AUT6
Find the sum of all possible values of z.
[2♣] Problem 5 (USAMO 1984/1). The product of two of the four roots of the quartic
equation x4 − 18x3 + kx2 + 200x − 1984 = 0 is −32. Determine k.
[3♣] Required Problem 6. The cubic x3 − 7x2 + 3x + 2 has irrational roots r, s, and
t, with r ≥ s ≥ t. There exists a unique set of rational numbers A, B, and C, such that
the cubic x3 + Ax2 + Bx + C has r + s as a root. What is A + B + C?
[3♣] Problem 7 (AIME II 2020, added by Benjamin Wang-Tie). Let P (x) = x2 − 3x − 7,
and let Q(x) and R(x) be two quadratic polynomials also with the coefficient of x2 equal
to 1. David computes each of the three sums P + Q, P + R, and Q + R and is surprised
to find that each pair of these sums has a common root, and these three common roots
are distinct. If Q(0) = 2, compute R(0).
[2♣] Problem 8 (Austria 2016/6, added by Abdullahil Kafi). Let a, b, c be three integers
for which the sum
ab ac bc
+
+
c
b
a
is an integer. Prove that each of the three numbers
ab
,
c
ac
,
b
bc
a
is an integer.
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Evan Chen《陳誼廷》 (OTIS, updated 2023-04-06)
ZEAC3666
Symmetric Polynomials
[3♣] Problem 9. Factor the polynomial
a(b − c)3 + b(c − a)3 + c(a − b)3 .
H1883820
[3♣] Problem 10. Let a, b, c be real numbers. Prove that
a3 + b3 + c3 = (a + b + c)3
BMCQ54
if and only if a5 + b5 + c5 = (a + b + c)5 .
[3♣] Problem 11. Let a, b, c, d be distinct real numbers such that
a+b+c+d=0
and
1 1 1 1
+ + + = 0.
a b
c d
Prove that two of the four numbers have sum zero.
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03AIMEII9
23HMMTT2
[3♣] Problem 12 (AIME II 2003, added by Lincoln Liu). Consider the polynomials
P (x) = x6 − x5 − x3 − x2 − x and Q(x) = x4 − x3 − x2 − 1. Given that z1 , z2 , z3 , and z4
are the roots of Q(x) = 0, find P (z1 ) + P (z2 ) + P (z3 ) + P (z4 ).
[5♣] Required Problem 13 (HMMT 2023/T2). Prove there don’t exist pairwise
distinct complex numbers a, b, c, and d such that
a3 − bcd = b3 − cda = c3 − dab = d3 − abc.
18CMIMCA9
[9♣] Problem 14 (CMIMC 2018 A9). Given the polynomial identity
(x2 − 3x + 1)1009 =
2018
X
ak xk
k=0
11SMTA7
NOTVIETA
14AMO1
13SMTA9
calculate the remainder when a20 + a21 + · · · + a22018 is divided by 2017.
[3♣] Required Problem 15 (Stanford Math Tournament 2011). Let P (x) be a polynomial of degree 2011 such that P (1) = 0, P (2) = 1, P (4) = 2, . . . , and P (22011 ) = 2011.
Find the coefficient of x1 in P .
[5♣] Problem 16. Suppose a, b, c are integers and the cubic polynomial x3 + ax2 + bx + c
√
has roots r ≤ s ≤ t. Show that rs + st + rt can be written as u ± v for some rational
numbers u and v.
[3♣] Problem 17 (USAMO 2014/1). Let a, b, c, d be real numbers such that b − d ≥ 5
and all zeros x1 , x2 , x3 , and x4 of the polynomial P (x) = x4 + ax3 + bx2 + cx + d are
real. Find the smallest value the product (x21 + 1)(x22 + 1)(x23 + 1)(x24 + 1) can take.
√ √ √
[3♣]√Required
Problem
√
√
√18 (Stanford
√
√ Math Tournament 2013). Let a = − 3+ 5+ 7,
b = 3 − 5 + 7, c = 3 + 5 − 7. Compute
a4
b4
c4
+
+
.
(a − b)(a − c) (b − c)(b − a) (c − a)(c − b)
Z5394300
[5♣] Problem 19 (Ritwin Narra). Prove that for any integer n 6= 1, if real numbers a,
b, c, d satisfy
a + b + c + d = an + bn + cn + dn = 0
then two of a, b, c, d sum to 0.
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Evan Chen《陳誼廷》 (OTIS, updated 2023-04-06)
12BLACKMOP
Symmetric Polynomials
[5♣] Required Problem 20 (Black MOP 2012). Let ABC be a triangle and let hA ,
hB , hC be the lengths of the altitudes from A, B, and C. Let a = BC, b = CA, c = AB.
Suppose that
p
p
p
p
p
p
a + hB + b + hC + c + hA = a + hC + b + hA + c + hB .
Prove that triangle ABC is isosceles.
Z40B5559
[5♣] Problem 21 (Newton’s sums). Prove the Newton’s sums theorem given in the
reading.
[9♣] Problem 22 (Z is a ring). Suppose that α and β are complex numbers and monic
polynomials P, Q ∈ Z[x] satisfy P (α) = Q(β) = 0.
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(a) Show that there is monic polynomial R ∈ Z[x] such that R(α + β) = 0.
20HMMTA9
(b) Show that there is a monic polynomial S ∈ Z[x] such that S(αβ) = 0.
[9♣] Problem 23 (HMMT 2020, added by Guanjie
Liu). Let P (x) = x2020 + x + 2. Let
2020
Q(x) be the monic polynomial of degree 2 whose roots are the pairwise products of
the roots of P (x). Let α satisfy P (α) = 4. Compute the sum of all possible values of
Q(α2 )2 .
[1♣] Mini Survey. Fill out feedback on the OTIS-WEB portal when submitting this
problem set. Any thoughts on problems (e.g. especially nice, instructive, easy, etc.) or
overall comments on the unit are welcome.
In addition, if you have any suggestions for problems to add, or want to write hints for
one problem you really liked, please do so in the ARCH system!
The maximum number of [♣] for this unit is [92♣], including the mini-survey.
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Evan Chen《陳誼廷》 (OTIS, updated 2023-04-06)
Symmetric Polynomials
§3 Solutions to the walkthroughs
§3.1 Solution 1.8, USAMO 1975/3
Consider the polynomial Q(x) = (x + 1)P (x) − x. It has degree n + 1 and has zeros at
x = 0, . . . , n. Hence we need to have
Q(x) = (x + 1)P (x) − x = cx(x − 1)(x − 2) . . . (x − n)
for some constant c. Plugging in x = −1 gives 1 = c · (−1)n+1 · (n + 1)!. Thus
Q(n + 1) = (−1)n+1 . Conclude
P (n + 1) =
(n + 1) + Q(n + 1)
n + 1 + (−1)n+1
=
.
n+2
n+2
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