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c 2007 The Author(s) and The IMO Compendium Group

Polynomial Equations
Duˇan Djuki´
s
c

Contents
1
2

1

Introduction
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1
2

Introduction

The title refers to determining polynomials in one or more variables (e
...
with real or complex
coefficients) which satisfy some given relation(s)
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Determine the polynomials P for which 16P (x2 ) = P (2x)2
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Plugging x = 0 in the given relation yields 16P (0) = P (0)2 , i
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P (0) = 0 or 16
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Then P (x) = xQ(x) for some polynomial Q and 16x2 Q(x2 ) =
4x2 Q(2x)2 , which reduces to 4Q(x2 ) = Q(2x)2
...
Hence, P (x) = 4 xR(x), with R satifying the same relation as P
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Putting P (x) = xQ(x) + 16 in the given relation we obtain 4xQ(x2 ) = xQ(2x)2 + 16Q(2x); hence Q(0) = 0, i
...
Q(x) = xQ1 (x) for
some polynomial Q1
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Thus Q(x) = x2 Q1 (x)
...
Then R
satisfies 4xn+1 R(x2 ) = 22n xn+1 R(2x)2 + 2n+4 R(2x), which implies that R(0) = 0,
a contradiction
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We conclude that P (x) = 16

n
1
4x

for some n ∈ N0
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We start by proving the following lemma (to be used frequently):
Lemma 1
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Proof
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The coefficient at x2n−1 is
2an an−1 , from which we get an−1 = 0
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Continuing in this manner we conclude that an−2k−1 = 0 for
k = 0, 1, 2,
...
e
...
△
Since P (x)2 = 16P (x2 /4) is a polynomial in x2 , we have P (x) = Q(x2 ) or P (x) = xQ(x2 )
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imo
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yu, www
...
com
case we similarly get 4Q(x2 ) = Q(4x)2
...
Proceeding in this
k
way we find that P (x) = xi S(x2 ) for each k ∈ N and some i ∈ {0, 1,
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Now it is
enough to take k with 2k > deg P and to conclude that S must be constant
...
A simple verification gives us the general solution P (x) = 16 1 x for
4
n ∈ N0
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A majority of problems of this type can be solved by one of the above two methods (although
some cannot, making math more interesting!)
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Find all polynomials P such that P (x)2 + P ( x )2 = P (x2 )P ( x2 )
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By the introducing lemma there exists a polynomial Q such that P (x) = Q(x2 )
1
1
or P (x) = xQ(x2 )
...
We conclude that P (x) = Q(x2 ),
where Q is also a solution of the considered polynomial equation
...

2
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Suppose there exist such polynomials
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Putting P (x) = c(x − a1 ) · · · (x − ak ) we get c(Q(x) − a1 ) · · · (Q(x) − ak ) =
(x − 1)(x − 2) · · · (x − 15)
...
, 15}
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Now,
investigating parity of the remaining (three or five) coefficients we conclude that each of them
has the equally many odd roots
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3
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Solution
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We have a2 − 2a− 2 = 0
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For x = 1 we have 2aP1 (1) = 8P1 (1), which (since a = 4) gives us P1 (1) = 0, i
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P1 (x) = (x−1)P2 (x), so P (x) = (x−1)2 P2 (x)+a
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Substituting in the initial relation and simplifying yields (x − 1)n Q(x)2 +
2aQ(x) = 2(2x + 2)n Q(2x2 − 1), giving us Q(1) = 0, a contradiction
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4
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Solution
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Fixing deg P = n and comparing coefficients of
both sides we deduce that the coefficients of polynomial P√must be rational
...
e
...
However, this is impossible because P (a) must be of
√
the form p + q 21 for some rational p, q for the coefficients of P are rational
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5
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Write f as f = P/Q with P and Q coprime polynomials and Q monic
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The condition of the problem
became P (x2 )/Q(x2 ) = P (x)2 /Q(x)2 − a
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Therefore Q(x) = xn for some n ∈ N
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Let P (x) = a0 + a1 x + · · · + am−1 xm−1 + xm
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This is only possible if a = 2 and 2m − n = 0, or a = 0
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Find all polynomials P satisfying P (x2 + 1) = P (x)2 + 1 for all x
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By the introducing lemma, there is a polynomial Q such that P (x) = Q(x2 + 1) or
P (x) = xQ(x2 + 1)
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Substituting x2 + 1 = y yields Q(y 2 + 1) = Q(y)2 + 1 and
yQ(y 2 + 1) = (y − 1)Q(y)2 + 1, respectively
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Setting y = 1 we obtain that Q(2) = 1
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We thus obtain an infinite sequence of points at which Q takes value 1, namely the sequence
given by a0 = 2 and an+1 = a2 + 1
...

n
It follows that if Q ≡ 1, then P (x) = Q(x2 + 1)
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7
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Solution
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Therefore P (x) = S(x2 ) for some polynomial S
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This is
equivalent to R(x − 1 ) = R( 1 − x), i
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R(y) ≡ R(−y), where R is the polynomial
2
2
such that S(x) = R(x − 1 )
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4
8
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Solution
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This implies
that P2 (x)/P1 (x) does not depend on x
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Analogously,
P4 = dP3 for some constant d
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Moreover, we see that P1 (x)P1 (y) depends only on
xy, i
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f (x) = P1 (x)P1 (n/x) is the same for all positive divisors x of natural number
n
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e
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It is easily verified that this
is possible only when P1 (x) = xn for some n
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Therefore m = n and c = d = 1, and finally m = n = 1
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9
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(IMO 2004
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imo
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4

Solution
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For every x ∈ R the triple (a, b, c) =
(6x, 3x, −2x) satisfies the condition ab + bc + ca = 0
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, n the
following equality holds:
3i + 5i + (−8)i − 2 · 7i ai = 0
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Then K(i) = 3i + 5i + (−8)i − 2 · 7i = 0
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Only for i = 2 and i = 4 do we have K(i) = 0
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It is easily verified that all
such P (x) satisfy the required condition
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(a) If a real polynomial P (x) satisfies P (x) ≥ 0 for all x, show that there exist real polynomials A(x) and B(x) such that P (x) = A(x)2 + B(x)2
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Solution
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It follows from the condition P (x) ≥ 0 for all x that all the αi are even, and from the condition
P (x) ≥ 0 for all x ≥ 0 that (∀i) either αi is even or ai < 0
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The
well-known formula (a2 + γb2 )(c2 + γd2 ) = (ac + γbd)2 + γ(ad − bc)2 now gives a required
representation for their product P (x)
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Prove that if the polynomials P and Q have a real root each and
P (1 + x + Q(x)2 ) = Q(1 + x + P (x)2 ),
then P ≡ Q
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Note that there exists x = a for which P (a)2 = Q(a)2
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Now P (b) = Q(b) for b = 1+a+P (a)2
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12
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Solution
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Then
P (P (x)) − Q(Q(x)) = [Q(P (x)) − Q(Q(x))] + R(P (x))
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Hence the degree of Q(P (x))− Q(Q(x))
is n2 − n + k
...

In the remaining case when R ≡ c is constant, the condition P (P (x)) = Q(Q(x)) gives us
Q(Q(x) + c) = Q(Q(x)) − c, so the equality Q(y + c) = Q(y) − c holds for infinitely many
y, implying Q(y + c) ≡ Q(y) − c
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Duˇan Djuki´ : Polynomial Equations
s
c

5

13
...
Show that

1
a

+

1
b

+

1
c

> 1
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We use the following auxilliary statement
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If A, B and C are pairwise coprime polynomials with A+ B = C, then the degree
of each of them is less than the number of different zeroes of the polynomial ABC
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Let
m

l

k

A(x) =
i=1

(x − pi )ai ,

B(x) =
i=1

(x − qi )bi ,

C(x) =
i=1

(x − ri )ci
...
Our statement now follows from the fact that A and B are
coprime
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Each of their degrees a deg P ,
deg
1
b deg Q, c deg R is less than deg P + deg Q + deg R and hence a > deg P +deg P
Q+deg R , etc
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Corollary
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14
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Two cells are called neighbors if they have a common side
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m
n+1
2
Solution
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We
assign to the i-th parallel the polynomial pi (x) = ai1 + ai2 x + · · · + aim xm−1 and define
p0 (x) = pn+1 (x) = 0
...
e
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This sequence of polynomials is entirely determined by term p1 (x)
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Consider the sequence of polynomials ri (x) given by r0 = 0, r1 = 1 and ri+1 = (1 − x −
xm−1 )ri − ri−1
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Polynomial p1 = 0 of
degree < m for which pn+1 = 0 exists if and only if rn+1 (x) and xm − 1 are not coprime,
i
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if and only if there exists ε such that εm = 1 and rn+1 (ε) = 0
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Let us write
c = 1−ε−εm−1 and denote by u1 , u2 the zeroes of polynomial x2 −cx+1
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The
of the above recurrent sequence is xi = 1
1
u1 − u2
latter case is clearly impossible
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imo
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...
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to un+1 = un+1 and hence to ω n+1 = 1, where u1 = u2 ω, which holds if and only if (∃u2 )
1
2
u2 ω = 1 and u2 (1 + ω) = c
...

¯
¯
2kπ
2kπ
Now if ω = cos n+1 + i sin n+1 and ε = cos 2lπ + i sin 2lπ , the above equality becomes the
m
m
desired one

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