Polynomials $p(x)$ and $q(x)$ such that $p(x) \cdot (x + 2) + q(x) \cdot (2x + 1) = 1$ 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 $x$?
 A: 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)
A: 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 $\,\Bbb Z\,\cong\, \Bbb Z[x]\bmod (x\!-\!n).$
