# Irreducible polynomials via composition

Does there exist a polynomial with integers coefficients $p(x)$ such that for any polynomial with integer coefficients $g(x)$, $p(g(x))$ is irreducible in $Q[x]$?

• Would this be true for $x^4+1$? Aug 31 '16 at 11:30
• @user336: do you mean irreducible over $\Bbb Q$? Aug 31 '16 at 11:44
• @Watson By Gauss's lemma that is equivalent.
– user157036
Aug 31 '16 at 11:47
• @user336-iactuallychosethis: you could have meant over $\Bbb R$… in which can this is false, if we assume $deg(p)>1$. Btw, you should demand your polynomial $p$ to have degree at least $2$. Aug 31 '16 at 11:48
• @Watson Good point, I'll add it.
– user157036
Aug 31 '16 at 11:52

Note that $p(f(x))-p(g(x))$ is divisible by $f(x)-g(x)$. So, letting $f(x)=p(x)+x$ and $g(x)=x$, you get that:

$$p(p(x)+x)-p(x)$$

is divisible by $p(x)$, so $p(p(x)+x)$ is divisible by $p(x)$.

So if $p(x)$ is not constant or linear, we see that $g(x)=p(x)+x$ solves this.

The case of linear $p$ is easy to prove separately.

This is pretty much the same as the proof for the theorem:

For any $f(x)\in \mathbb Z[x]$ with $\deg f>0$, there exists an $n\in\mathbb Z$ so that $f(n)$ is not prime.

This proof applies to $R[x]$ for any integral domain $R$. (It fails for non-integral domains, because nothing assures that $p(p(x)+x)$ has degree greater than $p(x)$. But it still works for non-integral commutative rings $R$ if $p$ is monic and of degree greater than $1$.)

For example, if $p(x)=x^4+1$, then \begin{align}p(p(x)+x) &= (x^4+x+1)^4+1 \\&= (x^4+1) (x^{12}+4 x^9+3 x^8+6 x^6+8 x^5+3 x^4+4 x^3+6 x^2+4 x+2)\end{align} (via Wolfram Alpha.)

You can get infinitely many cases by taking $g(x)=a(x)p(x)+x$ for any $a(x)$.

Or even more generally, letting $g(x)=a(x)p(h(x))+h(x)$ for any $a(x),h(x)$, then $p(g(x))$ is divisible by $p(h(x))$.