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Difference between revisions of "1975 Canadian MO Problems/Problem 8"

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{{Old CanadaMO box|num-b=7|after = Last Question|year=1975}}
 
{{Old CanadaMO box|num-b=7|after = Last Question|year=1975}}
  
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== Solution 1==
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== Solution==
  
 
Let <math>f(n)</math> be the degree of polynomial <math>n</math>. We begin by noting that <math>f(P(x)) = k</math>. This is because the degree of the LHS is <math>f(P(x))^{f(P(x))}</math> and the RHS is <math>f(P(x))^k</math>. Now we split <math>P(x)</math> into two cases.
 
Let <math>f(n)</math> be the degree of polynomial <math>n</math>. We begin by noting that <math>f(P(x)) = k</math>. This is because the degree of the LHS is <math>f(P(x))^{f(P(x))}</math> and the RHS is <math>f(P(x))^k</math>. Now we split <math>P(x)</math> into two cases.
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In the first case, <math>P(x)</math> is a constant. This means that <math>c = c^k \Longrightarrow c\in \{0,1\}</math> or <math>c\in \{0,1,-1\}</math> if <math>k</math> is even.
 
In the first case, <math>P(x)</math> is a constant. This means that <math>c = c^k \Longrightarrow c\in \{0,1\}</math> or <math>c\in \{0,1,-1\}</math> if <math>k</math> is even.
  
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In the second case, <math>P(x)</math> is nonconstant with coefficients of <math>a_1,a_2,\hdots a_{k+1}</math>. If we divide by <math>P(x)</math> on both sides, then we have that <math>a_1 + \frac{a_2}{P(x)} + \frac{a_3}{P(x)^2} \hdots \frac{a_{k+1}}{P(x)^{k+1}} = 1</math>. This can only be achieved if <math>a_2,a_3, \hdots a_{k+1} = 0</math>. This is because if we factor out a <math>\frac{1}{P(x)}</math>, then clearly these terms are not constant. Thus, <math>a_1 = 1</math> and our second solution is <math>P(x) = x^k</math>.  
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In the second case, <math>P(x)</math> is non-constant with coefficients of <math>a_1,a_2,\hdots a_{k+1}</math>. If we divide by <math>P(x)</math> on both sides, then we have that <math>a_1 + \frac{a_2}{P(x)} + \frac{a_3}{P(x)^2} \hdots \frac{a_{k+1}}{P(x)^{k+1}} = 1</math>. This can only be achieved if <math>a_2,a_3, \hdots a_{k+1} = 0</math>. This is because if we factor out a <math>\frac{1}{P(x)}</math>, then clearly these terms are not constant. Thus, <math>a_1 = 1</math> and our second solution is <math>P(x) = x^k</math>.  
  
 
~bigbrain123
 
~bigbrain123
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~minor edits by clever14710owl

Latest revision as of 19:01, 7 April 2024

Problem 8

Let $k$ be a positive integer. Find all polynomials \[P(x) = a_0+a_1x+\cdots+a_nx^n\] where the $a_i$ are real, which satisfy the equation \[P(P(x)) = \{P(x)\}^k\].

1975 Canadian MO (Problems)
Preceded by
Problem 7 		1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • Followed by
Last Question



Solution

Let $f(n)$ be the degree of polynomial $n$. We begin by noting that $f(P(x)) = k$. This is because the degree of the LHS is $f(P(x))^{f(P(x))}$ and the RHS is $f(P(x))^k$. Now we split $P(x)$ into two cases.

In the first case, $P(x)$ is a constant. This means that $c = c^k \Longrightarrow c\in \{0,1\}$ or $c\in \{0,1,-1\}$ if $k$ is even.

In the second case, $P(x)$ is non-constant with coefficients of $a_1,a_2,\hdots a_{k+1}$. If we divide by $P(x)$ on both sides, then we have that $a_1 + \frac{a_2}{P(x)} + \frac{a_3}{P(x)^2} \hdots \frac{a_{k+1}}{P(x)^{k+1}} = 1$. This can only be achieved if $a_2,a_3, \hdots a_{k+1} = 0$. This is because if we factor out a $\frac{1}{P(x)}$, then clearly these terms are not constant. Thus, $a_1 = 1$ and our second solution is $P(x) = x^k$.

~bigbrain123 ~minor edits by clever14710owl

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