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Is there a pair of polynomials $p(x)$ and $q(x)$, each with integer coefficients, such that

$p(x) \cdot (x + 2) + q(x) \cdot (2x + 1) = 1$

for every real number $x$?

  • 1
    What did you try? – 2017-01-22
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    P,Q such that work $$\forall x ?$$ – 2017-01-22

2 Answers 2

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No.

$3(p(1)+q(1)) = 1$ => $p(1) + q(1) = \frac{1}{3}$, but $p(1)$ and $q(1)$ is integer. So, there are no such p(x) and q(x)

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    It is simpler to evaluate at $\,x =-2.\,$ Moreover, this is a more natural way to proceed viewing it as Bezout equation for $\,\gcd(2x+1,x+2) = 1\,$ -- see my answer. – 2017-01-22
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Evaluating at $\,x= -2\,$ yields $\,3\mid 1,\,$ contradiction. Below we show how this inference is simply a special case of the relationship between Bezout gcd equations and modular inverses.

Notice that if $\,p,q,f\,$ are polynomials with integer coefficients, and $\,n\,$ is an integer then

$\qquad \begin{align} &1 = q(x) f(x) + p(x)(x\!-\!n) \\ \overset{\large x\ =\ n}\Longrightarrow\ &1 = q(n)f(n)\Rightarrow \color{#c00}{f(n) = \pm 1}\end{align}\ \ $ but for $\ \ \begin{align} &f(x) = 2x\!+\!1,\ n=-2\\ &\color{#c00}{f(n)} =f(-2)=\color{#c00}{-3\neq \pm1}\end{align}$

This is a function (polynomial) analog of the well-known number case of using Bezout equations to compute modular inverses, since the above Bezout equation for $\,\gcd(f(x),x\!-\!n)= 1\,$ implies that $\,f(x)\,$ is invertible mod $\,x\!-\!n.\,$ i.e. $\,f(x)\equiv \color{#c00}{f(n)}\,$ is invertible in $\,\Bbb Z\,\cong\, \Bbb Z[x]\bmod (x\!-\!n).$