## Equations in Polynomials: Problems and Solutions
Probem 1 By the introducing lemma there exists a
polynomial \( Q \) such that \( P(x)=Q(x^2) \) or \( P(x)=xQ(x^2) \). In the
former case \( Q(x^2)^2+Q(\frac1{x^2})=Q(x^4)Q(\frac1{x^4}) \) and
therefore \( Q(x)^2+Q(\frac1x)=Q(x^2)Q(\frac1{x^2}) \) (which is
precisely the relation fulfilled by \( P \)), whereas in the latter
case (similarly) \( xQ(x)^2+\frac1xQ(\frac1x)^2=Q(x^2)
Q(\frac1{x^2}) \) which is impossible for the left and right hand
side have odd and even degrees, respectively. We conclude that
\( P(x)=Q(x^2) \), where \( Q \) is also a solution of the considered
polynomial equation. Considering the solution of the least degree
we find that \( P \) must be constant.
Probem 2 Suppose there exist such polynomials. Then \( \deg
P\cdot\deg Q=15 \), so \( \deg P=k\in\{3,5\} \). Putting \( P(x)=c(x-a_1)
\cdots(x-a_k) \) we get \( c(Q(x)-a_1)\cdots(Q(x)-a_k)=(x-1)(x-2)
\cdots(x-15) \). Thus the roots of polynomial \( Q(x)-a_i \) are
distinct and comprise the set \( \{1,2,\dots,15\} \). All these
polynomials mutually differ at the last coefficient only. Now,
investigating parity of the remaining (three or five) coefficients
we conclude that each of them has the equally many odd roots.
This is impossible, since the total number of odd roots is 8, not
divisible by 3 or 5.
Probem 3 Denote \( P(1)=a \). We have \( a^2-2a-2=0 \).
Substituting \( P(x)=(x-1)P_1(x)+a \) in the initial relation and
simplifying yields \( (x-1)P_1(x)^2+2aP_1(x)=4(x+1)P_1(2x^2-1) \). For
\( x=1 \) we have \( 2aP_1(1)=8P_1(1) \), which (since \( a\neq4 \)) gives us
\( P_1(1)=0 \), i.e. \( P_1(x)=(x-1)P_2(x) \), so \( P(x)=(x-1)^2P_2(x)+a \).
Suppose that \( P(x)=(x-1)^nQ(x)+a \), where \( Q(1)\neq0 \). Substituting
in the initial relation and simplifying yields \( (x-1)^nQ(x)^2+
2aQ(x)=2(2x+2)^nQ(2x^2-1) \), giving us \( Q(1)=0 \), a contradiction.
It follows that \( P(x)=a \).
Probem 4 Suppose that \( P \) is not constant. Fixing \( \deg P=
n \) and comparing coefficients of both sides we deduce that the
coefficients of polynomial \( P \) must be rational. On the other
hand, setting \( x=a \) with \( a=a^2-4a+1 \), that is, \( a=\frac{5\pm
\sqrt{21}}2 \), we obtain \( P(a)=b \), where \( b^2-4b-1=0 \), i.e. \( b=2
\pm\sqrt5 \). However, this is impossible because \( P(a) \) must be of
the form \( p+q\sqrt{21} \) for some rational \( p,q \) for the
coefficients of \( P \) are rational. It follows that \( P(x) \) is
constant.
Probem 5 Write \( f \) as \( f=P/Q \) with \( P \) and \( Q \) coprime
polynomials and \( Q \) monic. By comparing leading coefficients we
obtain that \( P \) too is monic. The condition of the problem became
\( P(x^2)/Q(x^2)=P(x)^2/Q(x)^2-a \). Since \( P(x^2) \) and \( Q(x^2) \) are
coprime (if, to the contrary, they had a zero in common, then so
do \( P \) and \( Q \)), it follows that \( Q(x^2)=Q(x)^2 \). Therefore
\( Q(x)=x^n \) for some \( n\in\mathbb{N} \). Now we have
\( P(x^2)=P(x)^2-ax^{2n} \).
Let \( P(x)=a_0+a_1x+\cdots+a_{m-1}x^{m-1}+x^m \). Comparing coefficients of \( P(x)^2 \) and \( P(x^2) \) gives us \( a_{n-1}=\cdots= a_{2m-n+1}=0 \), \( a_{2m-n}=a/2 \), \( a_1=\cdots=a_{m-1}=0 \) and \( a_0=1 \). This is only possible if \( a=2 \) and \( 2m-n=0 \), or \( a=0 \).
Probem 6 By the introducing lemma, there is a polynomial
\( Q \) such that \( P(x)=Q(x^2+1) \) or \( P(x)=xQ(x^2+1) \). Then
\( Q((x^2+1)^2+1)=Q(x^2+1)^2-1 \) or \( (x^2+1)Q((x^2+1)^2+1)=
x^2Q(x^2+1)^2+1 \), respectively. Substituting \( x^2+1=y \) yields
\( Q(y^2+1)=Q(y)^2+1 \) and \( yQ(y^2+1)=(y-1)Q(y)^2+1 \), respectively.
Suppose that \( yQ(y^2+1)=(y-1)Q(y)^2+1 \). Setting \( y=1 \) we obtain that \( Q(2)=1 \). Note that, if \( a\neq0 \) and \( Q(a)=1 \), then also \( aQ(a^2+1)=(a-1)+1 \) and hence \( Q(a^2+1)=1 \). We thus obtain an infinite sequence of points at which \( Q \) takes value 1, namely the sequence given by \( a_0=2 \) and \( a_{n+1}=a_n^2+1 \). Therefore \( Q\equiv1 \). It follows that if \( Q\not\equiv 1 \), then \( P(x)=Q(x^2+1) \). Now we can easily list all solutions: these are the polynomials of the form \( T(T(\cdots(T(x))\cdots)) \), where \( T(x)=x^2+1 \).
Probem 7 It follows from the condition of the problem that
\( P(-\sin x)=P(\sin x) \), so \( P(-t)=P(t) \) for infinitely many \( t \);
hence the polynomials \( P(x) \) and \( P(-x) \) coincide. Therefore
\( P(x)=S(x^2) \) for some polynomial \( S \). Now \( S(\cos^2 x)=S(\sin^2
x) \) for all \( x \), so \( S(1-t)=S(t) \) for infinitely many \( t \), which
gives us \( S(x)\equiv S(1-x) \). This is equivalent to \( R(x-\frac12)=
R(\frac12-x) \), i.e. \( R(y)\equiv R(-y) \), where \( R \) is the
polynomial such that \( S(x)=R(x-\frac12) \). Now \( R(x)=T(x^2) \) for
some polynomial \( T \), and finally, \( P(x)=S(x^2)=R(x^2-\frac12)=
T(x^4-x^2+\frac14)=Q(x^4-x^2) \) for some polynomial \( Q \).
Probem 8 Clearly \( P_1(x)P_2(y)=P_2(x)P_1(y) \) for all
natural numbers \( x \) and \( y \). This implies that \( P_2(x)/P_1(x) \) does not depend
on \( x \). Hence \( P_2=cP_1 \) for some constant \( c \). Analogously,
\( P_4=dP_3 \) for some constant \( d \). Now we have \( cP_1(x)P_1(y)-dP_3
(z)P_3(t)=1 \) for all natural \( x,y,z,t \) with \( xy-zt=1 \). Moreover,
we see that \( P_1(x)P_1(y) \) depends only on \( xy \), i.e. \( f(x)=P_1(x)
P_1(n/x) \) is the same for all positive divisors \( x \) of natural
number \( n \). Since \( f(x) \) is a rational function and the number of
divisors \( x \) of \( n \) can be arbitrarily large, it follows that \( f \)
is constant in \( x \), i.e. a polynomial in \( n \). It is easily
verified that this is possible only when \( P_1(x)=x^n \) for some
\( n \). Similarly, \( P_3(x)=x^m \) for some \( m \) and \( c(xy)^n-d(zt)^m=1 \).
Therefore \( m=n \) and \( c=d=1 \), and finally \( m=n=1 \). So,
\( P_1(x)=P_2(x)=P_3(x)=P_4(x)=x \).
Probem 9 (IMO
2004.2) Let \( P(x)=a_0+a_1x+\dots+a_nx^n \). For every \( x\in
\mathbb{R} \) the triple \( (a,b,c)=(6x,3x,-2x) \) satisfies the
condition \( ab+bc+ca=0 \). Then the condition on \( P \) gives us \( P(3x)+
P(5x)+P(-8x)=2P(7x) \) for all \( x \), implying that for all \( i=0,1,2,
\dots,n \) the following equality holds:
\[ \left(3^i+5^i+(-8)^i-2\cdot 7^i\right)a_i=0.\]
Suppose that \( a_i\neq0 \). Then \( K(i)=3^i+5^i+(-8)^i-2\cdot 7^i=0 \).
But \( K(i) \) is negative for \( i \) odd and positive for \( i=0 \) or
\( i\geq6 \) even. Only for \( i=2 \) and \( i=4 \) do we have \( K(i)=0 \). It
follows that \( P(x)=a_2x^2+a_4x^4 \) for some real numbers \( a_2,a_4 \).
It is easily verified that all such \( P(x) \) satisfy
the required condition.
Probem 10 - (a) If a real polynomial \( P(x) \) satisfies \( P(x)\geq0 \) for all
\( x \), show that there exist real polynomials \( A(x) \) and \( B(x) \) such
that \( P(x)=A(x)^2+B(x)^2 \).
- (b) If a real polynomial \( P(x) \) satisfies \( P(x)\geq0 \) for all \( x\geq0 \), show that there exist real polynomials \( A(x) \) and \( B(x) \) such that \( P(x)=A(x)^2+xB(x)^2 \).
Polynomial \( P(x) \) can be written in the form
\[ P(x)=(x-a_1)^{\alpha_1}\cdots(x-a_k)^{\alpha_k}\cdot(x^2-b_1x+c_1)
\cdots(x^2-b_mx+c_m),\quad\quad\quad\quad\quad(\ast)\] where \( a_i,b_j,c_j \) are real
numbers such that \( a_i \) are distinct and the polynomials
\( x^2-b_ix+c_i \) have no real roots.
It follows from the condition \( P(x)\geq0 \) for all \( x \) that all the \( \alpha_i \) are even, and from the condition \( P(x)\geq0 \) for all \( x\geq0 \) that \( (\forall i) \) either \( \alpha_i \) is even or \( a_i< 0 \). This ensures that each linear or quadratic factor in \( (\ast) \) can be written in the required form \( A^2+B^2 \) and/or \( A^2+xB^2 \). The well-known formula \( (a^2+\gamma b^2)(c^2+\gamma d^2)=(ac+\gamma bd)^2+\gamma(ad-bc)^2 \) now gives a required representation for their product \( P(x) \).
Probem 11 Note that there exists \( x=a \) for which \( P(a)^2=
Q(a)^2 \). This follows from the fact that, if \( p \) and \( q \) are the
respective real roots of \( P \) and \( Q \), then \( P(p)^2-Q(p)^2\leq 0
\leq P(q)^2-Q(q)^2 \), and moreover \( P^2-Q^2 \) is continuous. Now
\( P(b)=Q(b) \) for \( b=1+a+P(a)^2 \). Taking \( a \) to be the largest real
number for which \( P(a)=Q(a) \) leads to an immediate contradiction.
Probem 12 Suppose that \( R=P-Q\neq0 \) and that \( 0< k\leq n-1 \)
is the degree of \( R(x) \). Then
\[ P(P(x))-Q(Q(x))=[Q(P(x))-Q(Q(x))]+R(P(x)).\] Putting \( Q(x)=
x^n+\cdots+a_1x+a_0 \) we have \( Q(P(x))-Q(Q(x))=[P(x)^n-Q(x)^n]+
\cdots+a_1[P(x)-Q(x)] \), where all summands but the first have
a degree at most \( n^2-n \), while the first summand equals \( R(x)
\cdot\left(P(x)^{n-1}+P(x)^{n-2}Q(x)+\cdots+Q(x)^{n-1}\right) \) and
therefore has the degree \( n^2-n+k \) with the leading coefficient
\( n \). Hence the degree of \( Q(P(x))-Q(Q(x)) \) is \( n^2-n+k \). The
degree of \( R(P(x)) \) is equal to \( kn< n^2-n+k \), from what we
conclude that the degree of the difference \( P(P(x))-Q(Q(x)) \) is
\( n^2-n+k \), a contradiction.
In the remaining case when \( R\equiv c \) 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 \( Q(y+c) \equiv Q(y)-c \). But this is only possible for \( c=0 \).
Probem 13 We use the following auxilliary statement.
Apply this statement on polynomials \( P^a,Q^b,R^c \). Each of their degrees \( a\deg P \), \( b\deg Q \), \( c\deg R \) is less than \( \deg P+\deg Q+\deg R \) and hence \( \frac1a> \frac{\deg P}{\deg P+\deg Q+\deg R} \), etc. Summing up yields the desired inequality.
Probem 14 Denote by \( a_{ij} \) the number in the intersection
of \( i \)-th parallel and \( j \)-th meridian. We assign to the \( i \)-th
parallel the polynomial \( p_i(x)=a_{i1}+a_{i2}x+\cdots+a_{im}
x^{m-1} \) and define \( p_0(x)=p_{n+1}(x)=0 \). The property that each
number equals the sum of its neighbors can be written as \( p_i(x)=
p_{i-1}(x)+p_{i+1}(x)+(x^{m-1}+x)p_i(x) \) modulo \( x^m-1 \), i.e.
\[ p_{i+1}(x)=(1-x-x^{m-1})p_i(x)-p_{i-1}(x)\quad\mbox{\rm(mod
\( x^m-1 \))}.\] This sequence of polynomials is entirely determined
by term \( p_1(x) \). The numbers \( a_{ij} \) can be written in the
required way if and only if a polynomial \( p_1(x)\neq0 \) of degree
less than \( m \) can be chosen so that \( p_{n+1}(x)=0 \).
Consider the sequence of polynomials \( r_i(x) \) given by \( r_0=0 \), \( r_1=1 \) and \( r_{i+1}=(1-x-x^{m-1})r_i-r_{i-1} \). Clearly, \( p_{n+1}(x)\equiv r_{n+1}(x)p_1(x) \) {\rm(mod \( x^m-1 \))}. Polynomial \( p_1\neq0 \) of degree \( < m \) for which \( p_{n+1}=0 \) exists if and only if \( r_{n+1}(x) \) and \( x^m-1 \) are not coprime, i.e. if and only if there exists \( \varepsilon \) such that \( \varepsilon^m=1 \) and \( r_{n+1}(\varepsilon)=0 \). Now consider the sequence \( (x_i) \) given by \( x_0=0 \), \( x_1=1 \) and \( x_{i+1}=(1-\varepsilon-\varepsilon^{m-1}) x_i-x_{i-1} \). Let us write \( c=1-\varepsilon-\varepsilon^{m-1} \) and denote by \( u_1,u_2 \) the zeroes of polynomial \( x^2-cx+1 \). The general term of the above recurrent sequence is \( x_i= \frac{u_1^i-u_2^i}{u_1-u_2} \) if \( u_1\neq u_2 \) and \( x_i= iu_1^i \) if \( u_1=u_2 \). The latter case is clearly impossible. In the former case (\( u_1\neq u_2 \)) equality \( x_{n+1}=0 \) is equivalent to \( u_1^{n+1}=u_2^{n+1} \) and hence to \( \omega^{n+1}=1 \), where \( u_1=u_2\omega \), which holds if and only if \( (\exists u_2) \) \( u_2^2\omega=1 \) and \( u_2(1+\omega)=c \). Therefore \( (1+\omega)^2=c^2 \omega \), so \[ 2+\omega+\bar\omega=(1-\varepsilon-\bar\varepsilon) ^2.\] Now if \( \omega=\cos\frac{2k\pi}{n+1}+i\sin\frac{2k\pi}{n+1} \) and \( \varepsilon=\cos\frac{2l\pi}m+i\sin\frac{2l\pi}m \), the above equality becomes the desired one. |

2005-2017 IMOmath.com | imomath"at"gmail.com | Math rendered by MathJax Home | Olympiads | Book | Training | IMO Results | Forum | Links | About | Contact us |