Is $x^j+x^k+2$ irreducible whenever $j+k$ is odd?

Question

Let $j,k$ be positive integers with $j>k$ and consider the polynomial

$$f(x)=x^j+x^k+2$$

I want to prove the conjecture :

$f(x)$ is irreducible in $\mathbb Q[x]$, whenever $j+k$ is odd. This is true for $j\le 300$ as I checked with PARI/GP.

If $f$ has real roots, they obviously must be negative and the absolute value of any root must be less than $2$ for $j>2$.

Moreover, $-1$ cannot be a root because of $f(-1)=2$, so $f(x)$ never can have a linear factor.

Can we use the bound of the absolute values of the roots and that the constant coefficient is prime to show that $f(x)$ must be irreducible in $\mathbb Q[x]$

Answer 1

Can we use the bound of the absolute values of the roots and that the constant coefficient is prime to show that $f(x)$ must be irreducible in $\mathbb Q[x]$

Yes, that is indeed exactly how we can prove the irreducibility of $x^j+x^k+2$, by more closely inspecting its (complex!) roots.

Firstly, no root $\alpha \in \mathbb{C}$ can satisfy $|\alpha|<1$, since then we would have $$2=|\alpha^j+\alpha^k|\leq|\alpha|^j+|\alpha|^k<1+1=2.$$ We also cannot have $|\alpha|=1$ with odd $j+k$, since then $2=|\alpha^k||\alpha^{j-k}+1|=|\alpha^{j-k}+1|$ implies $\alpha^{j-k}=1$ (for example by looking at the complex plane). However from $(\alpha^{j-k}+1)\alpha^k=-2$ we have $\alpha^k=-1$, a contradiction with $\alpha$ being an odd root of unity.

So we have $|\alpha|>1$, and the irreducibility now follows in a standard way: assuming a monic factor $p(x)$ with constant coefficient equal $\pm 1$ (we can choose such since $2$ is a prime), product of its roots $\alpha_i$ (which are also roots of the original polynomial) are given by $1=|p(0)|=\prod |\alpha_i|>1$, a contradiction.

Remark: When $j+k$ is even, there can be roots on the unit circle and we this reasoning would not work. Consider for example $\alpha=i$ in $x^6+x^2+2=(x^2+1)(x^4-x^2+2).$

Remark 2: It turns out the $j+k$ being odd is just a special case of a more general condition. Specifically for $j>k>0$ we have: $$ \bbox[#ffd,15px]{f(x)=x^j+x^k+2 \text{ is irreducible over } \mathbb{Q} \iff \nu_2(j) \neq \nu_2(k).} $$

Proof. Assume $\nu_2(j) = \nu_2(k)$, then $k=2^u(2a+1)$, $j=2^u(2b+1)$ for some integers $a,b,u\geq 0$. Now let $\alpha$ be a root of $g(x)=x^{2^u}+1$, it follows $\alpha^k=\alpha^{2^u(2a+1)}=(\alpha^{2^u})^{2a+1}=(-1)^{2a+1}=-1$, and similarly $\alpha^j=-1$. Thus $f(\alpha)=-1-1+2=0$. Since $g$ is irreducible (see for example Proving that $x^{2^n} + 1$ is irreducible in $\mathbb Q[x]$), $g$ divides $f$ and clearly $f\neq g$, hence $f$ is reducible.

For the opposite direction, assume $\nu_2(j) \neq \nu_2(k)$. By the previous argument it is enough to rule out existence of root $|\alpha|=1$. So assume there is such root $\alpha$, and as as before, we use $2=|\alpha^{j-k}+1|$ to deduce $\alpha^{j-k}=1$ and consequently $\alpha^k=-1$, similarly for $\alpha^j=-1$. But this means $\alpha=e^{\frac{2m+1}{k}\pi i}$ for some integer $m$. Then, $-1=\alpha^j=e^{\frac{2m+1}{k} j \pi i}$, so $\frac{2m+1}{k} j \pi = (2n+1) \pi $ for an integer $n$. However this simplifies to $(2m+1)j=(2n+1)k$, which implies $\nu_2(j) = \nu_2(k)$, a contradiction.

Answer 2

Tiniest bit of progress settling the case $j=k+1$.

By the usual business with Gauss' Lemma It suffices to show that there are no polynomials of positive degree $p(x),q(x)\in\Bbb{Z}[x]$ such that $$p(x)q(x)=f(x)\tag{1}$$ Because $f(0)=2$ we can, without loss of generality, assume that $p(0)=\pm1$ and $q(0)=\pm2$. Reducing $(1)$ modulo two gives us that $$ \overline{p}(x)\overline{q}(x)=x^j+x^k=x^k(x^{j-k}+1). $$ Our assumptions about the constant terms then allow us to deduce that $$ \begin{aligned} \overline{p}(x)&\mid x^{j-k}+1,\ \text{and}\\ x^k&\mid\overline{q}(x). \end{aligned} $$ In particular, it follows that $\deg p(x)\le j-k$.

The case $j=k+1$ can then be handled easily. We have seen that $p(x)$ is linear, so $f(x)$ has an odd integer root. The only alternatives are $x=\pm1$, and these were already excluded.