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Given a multivariate polynomial $f(x_1,\dots,x_n)$ with integer coefficients, how to find an integer $m$ (if it exists) such that $f(x_1,\dots,x_n) + m$ factors into polynomials of smaller degrees?

Are there any simple criteria to identify cases when such $m$ does not exist?

Is it possible that more than one suitable values of $m$ exist?

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    $\begingroup$ The polynomial $(x_1+\cdots+x_n)^2+m$ becomes factorisable whenever $m=-k^2$ for some integer $k$. $\endgroup$ Mar 3, 2013 at 6:39
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    $\begingroup$ About your first question - there need not be any simple criteria to say when such $m$ does not exist. Take e.g polynomial $x^2 + ny + m$. As long as $n \neq 0$, no value of $m$ will make this factorisable. Examples of this kind suggests that a simple criteria in terms of coefficients of the polynomial may not exist. $\endgroup$
    – User3568
    Mar 3, 2013 at 7:32
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    $\begingroup$ For $n\geq 2$, the polynomial $(x_1^2+\cdots+x_n^2)+m$ is never reducible. $\endgroup$
    – Name
    Mar 3, 2013 at 7:33
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    $\begingroup$ @Amit, on the contrary, such examples all follow from Newton polytope considerations. If the Newton polytope is not decomposable as a Minkowski sum, then such an $m$ does not exist. $\endgroup$ Mar 3, 2013 at 13:46
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    $\begingroup$ Hilbert's irreducibility theorem says that for almost all $m$ the polynomial $f(x_1, \ldots, x_n)+m$ is irreducible. Here almost all means the the number of integers $|m|\leq x$ for which it is irreducible is $O(\sqrt x)$. $\endgroup$ Mar 14, 2013 at 14:15

1 Answer 1

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Suppose that $f$ cannot be written as $f(x_1,x_2,\dots,x_n)=h(g(x_1,x_2,\dots,x_n))$ with $h\in\mathbb C[x]$ of degree $\ge2$ and $g\in\mathbb C[x_1,x_2,\dots,x_n]$. For $m\in\mathbb C$ let $a_m$ be the number of irreducible factors of $f(x_1,x_2,\dots,x_n)+m$. Then Ewa Cygan proved (see here for a review and reference) that $\sum_{m\in\mathbb C}(a_m-1)\le\deg f-1$. (The case $n=2$ was shown previously by Josef Stein, see here.)

So, as hinted in the comments, there can be infinitely many $m$ with $f(x_1,x_2,\dots,x_n)+m$ reducible. But in that case $f$ has a rather specific shape.

The question also asks to decide if there is at least one $m$ making $f+m$ reducible. Here is a sketch on how to find the $m$'s in case that there are only finitely many: Suppose that $f+m=uv$ with non-constant polynomials. Pick $\bar x\in\mathbb C^n$ with $u(\bar x)=v(\bar x)=0$. (There probably is no such $\bar x$, so one should better work in the projective completion.) Then the partial derivatives $\frac{\partial f}{\partial x_i}=\frac{\partial u}{\partial x_i}v+u\frac{\partial v}{\partial x_i}$ vanish in $\bar x$.

So the recipe is as follows: Determine all (usually finitely many) common roots $\bar x$ of the partial derivatives $\frac{\partial f}{\partial x_i}$, set $m=-f(\bar x)$, and check if $f(x_1,\dots,x_n)+m$ is reducible.

As said above, one should better work projectively in order to avoid missing solutions $m$ coming from singularities at infinity.

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  • $\begingroup$ Thanks for the reference. It's quite intuitive to expect that there are no so many (if any) suitable $m$'s in the general case. However I mostly wonder about the computational aspect -- is there a computationally efficient way to determine at least one suitable $m$ or establish that there are none? $\endgroup$ Aug 10, 2015 at 15:46
  • $\begingroup$ Can you please illustrate how to work projectivity on the example $f=x^2 y + 2 x y + x + y - 2$. Adding $m=3$ makes it reducible, but this cannot be found in the usual coordinates because the resulting system has no solutions. What is the theory for projective coordinates and how to apply it to this equation? $\endgroup$ Jan 25 at 10:20
  • $\begingroup$ @BogdanGrechuk OK, what I said in my answer does not quite work here. In fact, $f+m$ has a singularity at infinity for every $m$, so it does not help in finding $m$. (But of course, one can find the possibilities for $m$ in this example by noting that one factor has to have degree $0$ in $y$.) $\endgroup$ Jan 25 at 13:35
  • $\begingroup$ Yes, of course this example is easy and can be solved by other methods. I just tried to develop the simplest example when the system has no (non-projective) solutions, to see how "work projectively" (as you suggested) help in such examples. If it does not, then the general method for finding m remains an open question :( $\endgroup$ Jan 25 at 14:23

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