# When adding a constant makes a multivariate polynomial reducible?

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?

• The polynomial $(x_1+\cdots+x_n)^2+m$ becomes factorisable whenever $m=-k^2$ for some integer $k$. – Venkataramana Mar 3 '13 at 6:39
• 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. – User3568 Mar 3 '13 at 7:32
• For $n\geq 2$, the polynomial $(x_1^2+\cdots+x_n^2)+m$ is never reducible. – Name Mar 3 '13 at 7:33
• @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. – Gjergji Zaimi Mar 3 '13 at 13:46
• 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 $O(\sqrt x)$. – Lior Bary-Soroker Mar 14 '13 at 14:15

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.
• 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? – Max Alekseyev Aug 10 '15 at 15:46