I came across the following problem:



> 
> What are conditions such that the polynomial $x^2+1$ divides $p(y)+q(z)+ax+b=F(x,\,y,\,z)$,?
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> 
> 


Here $p$ and $q$ are also polynomials and $a$, $b$ are real numbers. The main difficulty is that $F(x,\, y, \,z)$ has three variables, and the idea of using roots cannot apply here. I am expecting a relation between $x$ and $y,z$, but I not able to find it.



What are conditions such that the polynomial $x^2+1$ divides $p(y)+q(z)+ax+b=F(x,\, y, \,z)$?
Suppose $(x^2+1)$ divides $F(x,y,z)$, that is, $F(x,y,z) = (x^2+1)G(x,y,z)$ for some polynomial $G(x,y,z)$.
Then, setting $x = \pm \sqrt{-1}$, we see that
$$
F( \pm i, y,z) = 0.
$$
In your particular case, we must have that
$$
p(y)+q(z)\pm ai + b =0,
$$
Taking the difference of these two equations, we see that $2ai=0$, so $a=0$.
Hence, $F$ does not depend on $x$, so this is only possible if $G(x,y,z) \equiv 0$, so $F(x,y,z) \equiv 0$.


An alternative approach is to just compare coefficients. It is impossible for a non-zero polynomial to be a multiple of $(x^2+1)$ while its degree (in $x$) is $1$.


