Can we prove those conjectures about a family of polynomials?

Question

Let $\ a,b\ $ integers with $\ 0<a<b\ $.

Consider the following family of polynomials :

$$f(x)=x^b+x^a+1$$ $$g(x)=x^b+x^a-1$$ $$h(x)=x^b-x^a+1$$ $$i(x)=x^b-x^a-1$$

The problem is about which of those polynomials can be simultaneously reducible.

The following conjecture holds for every pair $\ (a,b)\ $ with $\ b\le 70\ $

• If $\ f\ $ is irreducible over $\ \mathbb Z[x]\ $ , then also $\ g,h\ $ and $\ i\ $.
• At least two of the polynomials are simultaneously irreducible over $\ \mathbb Z[x]\ $.

Can we prove those conjectures ?

It is clear that none of the polynomials have a linear factor. I am not sure which method could work : Substitution of $\ x-k\ $ for some suitable $\ k\ $ or reduction modulo some prime number $\ p\ $. Ideas , a reference and also the extension of the conjecture to higher limits are welcome.

Answer 1

For $b = 2a$ we can use Corollary 6 in Keith Conrad's Irreducibility of $x^n − x − 1$ to deduce irreducibility of both $x^b \pm x^a -1$, your $g(x)$ and $i(x)$. This settles your second claim. After the corollary it is conjectured that $f(x)$ is irreducible iff $a=3^j$ and $h(x)$ is irreducible iff $a=2^i 3^j$. First one is not hard to prove, I am not sure about the second one yet (it might be open). These imply your first claim.

For $b \neq 2a$, there is a complete irreducibility characterization of $x^b\pm x^a \pm 1$ by W. Ljunggren in On the irreducibility of certain trinomials and quadrinomials, Theorem 3 (see https://www.mscand.dk/article/view/10593 ). Technically the statement includes also the case $b=2a$, but there seems to be an error - for example for $x^4-x^2+1$ it claims reducibility yet it is irreducible. So in the rest of the post I assume the theorem is correct for $b \neq 2a$ (it was published and reviewed after all ...). Here is a restated version of the theorem:

Theorem. Let $f(x)=x^n+ \varepsilon x^m + \varepsilon'$ with $\varepsilon=\pm 1$, $\varepsilon'=\pm 1$ and $n > 2m$. Put $n=n_1d$, $m=m_1d$ where $\gcd(n_1,m_1)=1$. Then $f(x)$ is reducible if and only if $n_1 + m_1 \equiv 0 \pmod 3$ and

• $2 \nmid n_1m_1$, $\varepsilon=1$ or
• $2\mid n_1$, $\varepsilon'=1$ or
• $2\mid m_1$, $\varepsilon=\varepsilon'$.

In such case $f(x)$ is a product of $x^{2d}+\varepsilon^m \varepsilon'^n x^d+1$ and second irreducible polynomial.

Proof. In the linked article.

Note: The condition $n > 2m$ is not really a limitation: for $n<2m$ we can take the reciprocal polynomial $\overline{f}(x)=\varepsilon'x^m+\varepsilon x^{m-n}+1$, which after suitable multiplication by $-1$ (if needed) satisfies the conditions of the theorem.

Let's first assume $b > 2a$. Put $a_1=a/\gcd(a,b)$ and $b_1=b/\gcd(a,b)$, then by the above theorem for any of $f(x),g(x),h(x),i(x)$ to be reducible it is necessary $a_1 + b_1 \equiv 0 \pmod 3$. Under this assumption:

1. $f(x)$ is reducible.

2. $g(x)$ is reducible iff $2 \nmid a_1b_1$.

3. $h(x)$ is reducible iff $2 \mid b_1$.

4. $i(x)$ is reducible iff $2 \mid a_1$.

If $f(x)$ is irreducible, we must have $a_1 + b_1 \not\equiv 0 \pmod 3$ and so all the other polynomials are irreducible as well, hence your first claim. Moreover the conditions $2 \nmid a_1b_1$, $2 \mid a_1$ and $2 \mid b_1$ are pairwise mutually exclusive (remember $a_1,b_1$ are coprime), so at most one of $g(x), h(x), i(x)$ is reducible. Together with $f(x)$ reducible, we have at most two of the four polynomials reducible, hence your second claim.

Now you can do the same for $b<2a$, replacing the four polynomials with their reciprocal (e.g. $f(x)$ with $x^b+x^{b-a}+1$), multiply by $-1$ (if needed) to make the leading coefficient $1$ and apply the same method as above.