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Difference between revisions of "2013 Canadian MO Problems/Problem 1"
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Note that since <math>c_0</math>, <math>c_1</math>, and <math>c_2</math> are not present in the expression before <math>x^2</math>, they can be anything and the coefficient in front of <math>x^2</math> is still zero.
 
Note that since <math>c_0</math>, <math>c_1</math>, and <math>c_2</math> are not present in the expression before <math>x^2</math>, they can be anything and the coefficient in front of <math>x^2</math> is still zero.
  
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So now we just need to find <math>c_1</math> and <math>c_2</math>
+
So now we just need to find <math>c_1</math> and <math>c_2</math>, for that we look at the coefficient in front of <math>x</math> in <math>F(x)</math>:
−
 
 
−
So, we look at the coefficient in front of <math>x</math> in <math>F(x)</math>:
 
  
 
<math>\left( c_1-c_0+\sum_{i=1}^{n}(-1)^{i-1}\binom{i}{i-1}c_i+\sum_{i=0}^{n}(-1)^{i}\binom{i}{i}c_i \right)x</math>
 
<math>\left( c_1-c_0+\sum_{i=1}^{n}(-1)^{i-1}\binom{i}{i-1}c_i+\sum_{i=0}^{n}(-1)^{i}\binom{i}{i}c_i \right)x</math>

Revision as of 00:16, 27 November 2023

Problem

Determine all polynomials $P(x)$ with real coefficients such that \[(x+1)P(x-1)-(x-1)P(x)\] is a constant polynomial.

Solution

Let $F(x)=(x+1)P(x-1)-(x-1)P(x)$

$P(x)=\sum_{i=0}^{n}c_ix^i$

$F(x)=(x+1)\sum_{i=0}^{n}(x-1)^ic_i-(x-1)\sum_{i=0}^{n}c_ix^i$

$F(x)=\sum_{i=0}^{n}x(x-1)^ic_i+\sum_{i=0}^{n}(x-1)^ic_i-\sum_{i=0}^{n}c_ix^{i+1}+\sum_{i=0}^{n}c_ix^i$

$\sum_{i=0}^{n}(x-1)^ic_i=\sum_{j=0}^{n}\left( \sum_{i=j}^{n}(-1)^{i-j}\binom{i}{i-j}c_i \right)x^j$

$\sum_{i=0}^{n}x(x-1)^ic_i=\sum_{j=0}^{n}\left( \sum_{i=j}^{n}(-1)^{i-j}\binom{i}{i-j}c_i \right)x^{j+1}$

$F(x)=\sum_{j=0}^{n}\left( \sum_{i=j}^{n}(-1)^{i-j}\binom{i}{i-j}c_i \right)x^j+\sum_{j=0}^{n}\left( \sum_{i=j}^{n}(-1)^{i-j}\binom{i}{i-j}c_i \right)x^{j+1}-\sum_{i=0}^{n}c_ix^{i+1}+\sum_{i=0}^{n}c_ix^i$

In order for the new polynomial $F(x)$ to be a constant, all the coefficients in front of $x^i$ for $i>1$ need to be zero.

So we start by looking at the coefficient in front of $x^2$:

$\left( c_2-c_1+\sum_{i=2}^{n}(-1)^{i-2}\binom{i}{i-2}c_i+\sum_{i=1}^{n}(-1)^{i-1}\binom{i}{i-1}c_i \right)x^2$

Since $\sum_{i=1}^{n}(-1)^{i-1}\binom{i}{i-1}c_i=c_1+\sum_{i=2}^{n}(-1)^{i-1}\binom{i}{i-1}c_i$,

$\left( c_2-c_1+c_1+\sum_{i=2}^{n}(-1)^{i-2}\binom{i}{i-2}c_i+\sum_{i=2}^{n}(-1)^{i-1}\binom{i}{i-1}c_i \right)x^2$

$\left( c_2+\sum_{i=2}^{n}\left((-1)^{i-2}\binom{i}{i-2}+(-1)^{i-1}\binom{i}{i-1} \right)c_i\right)x^2$

We then evaluate the term of the sum when $i=2$:

$\left( c_2+(1-2)c_2+\sum_{i=3}^{n}\left((-1)^{i-2}\binom{i}{i-2}+(-1)^{i-1}\binom{i}{i-1} \right)c_i\right)x^2$

$\left(\sum_{i=3}^{n}\left((-1)^{i-2}\binom{i}{i-2}+(-1)^{i-1}\binom{i}{i-1} \right)c_i\right)x^2=0$

Therefore all coefficients $c_i$ for $i \ge 3$ need to be zero so that the coefficient in front of $x^2$ is zero.

That is, $c_3=c_4=c_5=\cdots =c_n=0$.

Note that since $c_0$, $c_1$, and $c_2$ are not present in the expression before $x^2$, they can be anything and the coefficient in front of $x^2$ is still zero.

So now we just need to find $c_1$ and $c_2$, for that we look at the coefficient in front of $x$ in $F(x)$:

$\left( c_1-c_0+\sum_{i=1}^{n}(-1)^{i-1}\binom{i}{i-1}c_i+\sum_{i=0}^{n}(-1)^{i}\binom{i}{i}c_i \right)x$

Since $c_i$=0 for $i \ge 3$:

$\left( c_1-c_0+\sum_{i=1}^{2}(-1)^{i-1}\binom{i}{i-1}c_i+\sum_{i=0}^{2}(-1)^{i}\binom{i}{i}c_i \right)x$

$\left( c_1-c_0+\binom{1}{0}c_1-\binom{2}{1}c_2+c_0-c_1+c_2\right)x$

$\left( c_1-c_0+c_1-2c_2+c_0-c_1+c_2\right)x$

$\left(c_1-c_2\right)x$

Therefore $c_1-c_2=0$, thus $c_1=c_2$ satisfies the condition for $F(x)$ to be a constant polynomial.

So we can set $c_1=c_2=A$ and $c_0=B$, and all the polynomials $P(x)$ will be in the form:

$P(x)=Ax^2+Ax+B$ where $A,B \in \mathbb{R}$

~Tomas Diaz. orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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