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

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 $\mathbb Q[x]$

• The nonexistence of roots is not enough to show irreducibility in $\mathbb{Q}[x]$, except when the degree is small. Sep 9, 2017 at 12:50
• Have you tried the standard trick of a shift and applying Eisenstein? Sep 9, 2017 at 13:01
• @MichaelBurr I have never claimed this, but to avoid down-/closevotes I decided to show some thoughts :) And I did not try Eisenstein, but I doubt it is successful here. Sep 9, 2017 at 13:35

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.

• Very sorry about having missed your solution for a long time. Good stuff! May 28, 2021 at 15:21
• @JyrkiLahtonen Thank you, I thought about it some more since I was wondering what happens in case of $j+k$ even, and turns out there is a simple characterization for irreducibility of which $j+k$ odd is just a special case, so I added that. Unless I made a mistake somewhere of course :)
– Sil
Jun 8, 2021 at 20:38

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.