For prime number $q$ the polynomial $P(x)$ with integer coefficients is said to be factorable if there exist non-constant polynomials $f_q,g_q$ with integer coefficients such that all of the coefficients of the polynomial $Q(x)=P(x)-f_q(x)g_q(x)$ are dividable by $q$ ; and we write: $$P(x)\equiv f_q(x)g_q(x)\pmod{q}$$For example the polynomials $2x^3+2,x^2+1,x^3+1$ can be factored modulo $2,3,p$ in the following way: $$\left\{\begin{array}{lll} X^2+1\equiv (x+1)(-x+1)\pmod{2}\\ 2x^3+2\equiv (2x-1)^3\pmod{3}\\ X^3+1\equiv (x+1)(x^2-x+1) \end{array}\right.$$ Also the polynomial $x^2-2$ is not factorable modulo $p=8k\pm 3$. a) Find all prime numbers $p$ such that the polynomial $P(x)$ is factorable modulo $p$: $$P(x)=x^4-2x^3+3x^2-2x-5$$b) Does there exist irreducible polynomial $P(x)$ in $\mathbb{Z}[x]$ with integer coefficients such that for each prime number $p$ , it is factorable modulo $p$?
Problem
Source: Iran third round 2017 number theory , final exam
Tags: number theory, polynomial, algebra
01.09.2017 12:30
This is part (a) (it was changed): prove that there exists infinitely many prime numbers $p$ such that $P(x)$ is factorable modula $p$: $$P(x)=x^4-2x^3+3x^2-2x-5$$
01.09.2017 12:34
(a) By Schur, there exists infinite many prime $p$ dividing $P(r)$ for some positive integer $r$. Therefore in $\mathbb{F}_p$, $(X-r)$ is a factor of $P(X)$ and we are done.
01.09.2017 12:42
(b)(This is not my solution, but it's really nice!) yes,$P(x)=x^6+3$ is irreducible over $Z$ but it is factorable modula $p$: if $p=3k+1$: we can write $-3 \equiv y^2(mod p)$ since $\left(\frac{-3}{p}\right) =1$ so $x^6+3 \equiv x^6-y^2 \equiv (x^3-y)(x^3+y) (modp)$ if $p=3k+2$:we can write $3 \equiv y^3 (mod p)$ since $(3)^{\frac {p-1}{gcd(3,p-1)}} \equiv 1(mod p)$ so $x^6+3 \equiv x^6+y^3 \equiv (x^2+y)(x^4+x^2y+y^2) (mod p)$ if $p=3$: $x^6+3 \equiv x^6 (mod 3)$
12.09.2018 16:02
For the second part $x^4+6x^2+1$ also works.It is worth mentioning that the second part is quite well-known.
03.12.2023 05:15
In my opinion, $x^4+1$ is even more classical for the second part. (Consider $-2$, $2$, $-1$ as quadratic residues mod $p$ separately.)