0
$\begingroup$

I came across the following problem:

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

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.

Improve this question
$\endgroup$
1
  • $\begingroup$ It should be the case that F(i,y,z) = F(-i,y,z) = 0, where i is the square root of -1. But, this seems to imply that $a=0$, and thus, your polynomial must be identically 0. Let me make that into an answer. $\endgroup$
    – Per Alexandersson
    Commented Feb 20, 2021 at 11:18

1 Answer 1

Reset to default
2
$\begingroup$

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 $1$.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ I think your second approach is preferable, since it also works in characteristic 2. And indeed it will even work for $R[x,y,z]$ where $R$ is an arbitrary ring, since $x^2+1$ is monic, so the $x$ degree of $(x^2+1)G(x,y,z)$ is always at least 2 unless $G$ is identically 0. $\endgroup$
    – Joe Silverman
    Commented Feb 20, 2021 at 12:16

Not the answer you're looking for? Browse other questions tagged .