2 Preliminaries

Throughout this article, we assume that $p$ is an odd prime and that

(2.1) $$ \begin{align} {\mathcal F}_{p}(x):=x^{2p}+2x^{p}+2. \end{align} $$

Note that ${\mathcal F}_p(x)$ is irreducible over ${\mathbb Q}$ since it is $2$ -Eisenstein. From a theorem due to Swan [Reference Swan6],

$$ \begin{align*} \Delta({\mathcal F}_{p})=-2^{3p-1}p^{2p}. \end{align*} $$

From (1.1) and (1.2),

(2.2) $$ \begin{align} {\mathcal{F}}_p(x) \ \text{is monogenic}\quad \text{if and only if} \quad \gcd(2p,\mathrm{index}({\mathcal{F}}_p))=1. \end{align} $$

One method that can be used, for any monic irreducible polynomial, to determine explicit conditions (more so than (2.2)) on the prime divisors of the polynomial discriminant to guarantee monogenicity is known as Dedekind’s index criterion [Reference Cohen1]. There is a more algorithmic method that is derived from Dedekind’s theorem and designed explicitly for trinomials. This streamlined version of Dedekind’s theorem is due to Jakhar et al. [Reference Jakhar, Khanduja and Sangwan3, Theorem 1.1], and requires that only one out of a list of five conditions be checked for each prime divisor $p$ of the discriminant of the trinomial to determine whether $p$ divides the index of the trinomial. Moreover, a careful analysis of [Reference Jakhar, Khanduja and Sangwan3, Theorem 1.1] reveals that we only need to apply the fourth condition of that theorem in our situation, which yields the following very succinct result.

Theorem 2.1. The trinomial ${\mathcal F}_p(x)$ is monogenic if and only if the polynomials

$$ \begin{align*} H_1(x):=x^{2}+2x+2 \quad \text{and}\quad H_2(x):=\dfrac{2x^{p}+2-(2x+2)^{p}}{p} \end{align*} $$

are coprime in ${\mathbb F}_p[x]$ .

Since

(2.3) $$ \begin{align} H_2(x)=\dfrac{2x^{p}+2-2^p(x+1)^{p}}{p} =\bigg(\frac{2-2^p}{p}\bigg)\, x^p+\frac{2-2^p}{p}-2^p\sum_{j=1}^{p-1}\frac{\binom{p}{j}}{p}\, x^j, \end{align} $$

we see indeed that $H_2(x)\in {\mathbb Z}[x]$ , by Fermat’s little theorem and the fact that $p$ divides the binomial coefficient $\binom {p}{j}$ for all $j$ with $1\le j\le p-1$ .

3 Proof of Theorem 1.1

From Theorem 2.1, we must determine exactly when $H_1(x)$ and $H_2(x)$ are coprime in ${\mathbb F}_p[x]$ . Observe from (2.3) that

$$ \begin{align*}\deg(H_2)\ge p-1\ge 2=\deg(H_1)\end{align*} $$

in ${\mathbb F}_p[x]$ . Thus, our general strategy is to determine the conditions under which $pH_2(\theta )\equiv 0 \pmod {p^2}$ for the zeros $\theta $ of $H_1(x)$ , which will tell us precisely when $H_1(x)$ and $H_2(x)$ are not coprime in ${\mathbb F}_p[x]$ , and consequently, when ${\mathcal F}_p(x)$ is not monogenic, by Theorem 2.1. We split the proof into the two cases $p\equiv \pm 1\pmod {4}$ .

Suppose first that $p\equiv -1 \pmod {4}$ . Then $H_1(x)$ is irreducible in ${\mathbb F}_p[x]$ since $-1$ is not a square in ${\mathbb F}_p$ . Hence, if $H_1(x)$ and $H_2(x)$ are not coprime in ${\mathbb F}_p[x]$ , it must be that $H_1(x)$ is an irreducible factor of $H_2(x)$ , so that every zero of $H_1(x)$ is a zero of $H_2(x)$ . Therefore, it is enough to examine $pH_2(\theta )$ modulo $p^2$ for any zero $\theta $ of $H_1(x)$ . Observe that $\alpha :=-1+i$ is a zero of $H_1(x)$ , where $i^2=-1$ and $i\not \in {\mathbb F}_p$ . Thus, to establish the theorem in this case, we show that

(3.1) $$ \begin{align} pH_2(\alpha)\equiv 0 \pmod{p^2} \quad \text{if and only if} \quad 2^{p-1}\equiv 1 \pmod{p^2}. \end{align} $$

Since $(\alpha +1)^p=i^p=-i$ , from the definition of $H_2(x)$ ,

(3.2) $$ \begin{align} pH_2(\alpha)=2\alpha^p+2-2^p(\alpha+1)^p=2\alpha^p+2+2^pi. \end{align} $$

A straightforward induction argument shows that

$$ \begin{align*}\alpha^j=(-1)^{{(j-3)}/{4}}2^{{(j-1)}/{2}}(1+i) \quad \text{if } j\equiv 3 \pmod{4}.\end{align*} $$

Thus,

$$ \begin{align*}\alpha^p=(-1)^{{(p-3)}/{4}}2^{{(p-1)}/{2}}(1+i),\end{align*} $$

which yields the following rearrangement of $pH_2(\alpha )$ from (3.2):

(3.3) $$ \begin{align} pH_2(\alpha)&=2((-1)^{{(p-3)}/{4}}2^{{(p-1)}/{2}}(1+i))+2+2^pi \nonumber \\ &=2+(-1)^{{(p-3)}/{4}}2^{{(p+1)}/{2}}+(2^p+(-1)^{{(p-3)}/{4}}2^{{(p+1)}/{2}})i. \end{align} $$

Consequently, from (3.3), it follows that $pH_2(\alpha )\equiv 0 \pmod {p^2}$ if and only if both the following congruences hold:

(3.4) $$ \begin{align} \begin{aligned} 2+(-1)^{{(p-3)}/{4}}2^{{(p+1)}/{2}} &\equiv 0 \pmod{p^2}, \\ 2^p+(-1)^{{(p-3)}/{4}}2^{{(p+1)}/{2}} &\equiv 0 \pmod{p^2}. \end{aligned} \end{align} $$

To prove (3.1), suppose first that $pH_2(\alpha )\equiv 0 \pmod {p^2}$ . From the first congruence in (3.4),

(3.5) $$ \begin{align} (-1)^{{(p-3)}/{4}}2^{{(p-1)}/{2}}\equiv -1 \pmod{p^2}. \end{align} $$

Squaring both sides of (3.5) yields $2^{p-1}\equiv 1 \pmod {p^2}$ .

Conversely, suppose that $2^{p-1}\equiv 1 \pmod {p^2}$ . Then

(3.6) $$ \begin{align} 2^{p-1}-1\equiv (2^{{(p-1)}/{2}}-1)(2^{{(p-1)}/{2}}+1)\equiv 0 \pmod{p^2}. \end{align} $$

From (3.6) we conclude that

$$ \begin{align*}2^{{(p-1)}/{2}}\equiv \pm 1 \pmod{p^2},\end{align*} $$

since $p$ divides one and only one of the factors in the factorisation (3.6). We claim that

(3.7) $$ \begin{align} 2^{{(p-1)}/{2}}\equiv (-1)^{{(p+1)/}{4}} \pmod{p^2}. \end{align} $$

Suppose first that $2^{{(p-1)}/{2}}\equiv 1 \pmod {p^2}$ , so that $2^{{(p-1)}/{2}}\equiv 1 \pmod {p}$ . By Euler’s criterion and the quadratic character of 2,

$$ \begin{align*}2^{{(p-1)}/{2}}\equiv \bigg(\frac{2}{p}\bigg)\equiv (-1)^{{(p^2-1)}/{8}} \pmod{p}.\end{align*} $$

Since $p\equiv 3\pmod {4}$ , it follows that $p\equiv 7 \pmod {8}$ and thus,

$$ \begin{align*}(-1)^{{(p^2-1)}/{8}}=(-1)^{{(p+1)}/{4}}=1.\end{align*} $$

Similarly, if $2^{{(p-1)}/{2}}\equiv -1 \pmod {p^2}$ , then $p\equiv 3 \pmod {8}$ , so that

$$ \begin{align*}(-1)^{{(p^2-1)}/{8}}=(-1)^{{(p+1)}/{4}}=-1.\end{align*} $$

Therefore, the claim (3.7) is established. Since $2^p\equiv 2\pmod {p^2}$ and $p\equiv 3 \pmod {4}$ , it follows from (3.3) and (3.7) that

$$ \begin{align*} pH_2(\alpha)&\equiv 2(1+(-1)^{{(p-3)}/{4}}2^{{(p-1)}/{2}})(1+i)\\ &\equiv 2(1+(-1)^{{(p-3)}/{4}}(-1)^{{(p+1)}/{4}})(1+i)\\ &\equiv 2(1+(-1)^{{(p-1)}/{2}})(1+i)\\ &\equiv 0 \pmod{p^2}, \end{align*} $$

which finishes the proof of (3.1), and completes the proof of the theorem when ${p\equiv 3 \pmod {4}}$ .

Suppose next that $p\equiv 1 \pmod {4}$ . Then $H_1(x)$ is reducible in ${\mathbb F}_p[x]$ since $-1$ is a square in ${\mathbb F}_p$ . Let $\alpha =-1+i\in {\mathbb F}_p$ be a zero of $H_1(x)$ , where $i$ and $\alpha $ are such that $1\le \alpha \le p-3$ and $2\le i\le p-2$ with $i^2\equiv -1 \pmod {p}$ . Then $\beta =p-1-i\in {\mathbb F}_p[x]$ is the other zero of $H_1(x)$ , where $1\le \beta \le p-3$ . The strategy here, as in the previous case, is to determine conditions under which $pH_2(\theta )\equiv 0 \pmod {p^2}$ , where $\theta \in \{\alpha ,\beta \}$ . It is conceivable that exactly one of the two zeros $\theta $ , and not the other, could satisfy the congruence $pH_2(\theta )\equiv 0 \pmod {p^2}$ . However, it turns out that

(3.8) $$ \begin{align} pH_2(\alpha)\equiv 0 \pmod{p^2} \quad \text{if and only if} \quad pH_2(\beta)\equiv 0 \pmod{p^2}, \end{align} $$

which shows that considering the single zero $\alpha $ will suffice. We postpone the proof of (3.8) to focus first on $\alpha $ , and show that

(3.9) $$ \begin{align} pH_2(\alpha)\equiv 0 \pmod{p^2} \quad \text{if and only if} \quad 2^{p-1}\equiv 1 \pmod{p^2}. \end{align} $$

Since $H_1(\alpha )\equiv 0 \pmod {p}$ , we see that $\alpha ^2\equiv -2(\alpha +1) \pmod {p}$ , which implies

(3.10) $$ \begin{align} \alpha^{2p}\equiv -2^p(\alpha+1)^p\equiv -2^pi^p \pmod{p^2}. \end{align} $$

Therefore, from the definition of $H_2(x)$ , (2.1) and (3.10), it follows that

(3.11) $$ \begin{align} pH_2(\alpha)&=2\alpha^p+2-2^p(\alpha+1)^p \nonumber\\ &=2\alpha^p+2-2^pi^p \nonumber\\ &\equiv 2\alpha^p+2+\alpha^{2p} \equiv {\mathcal F}_p(\alpha) \pmod{p^2}. \end{align} $$

To prove (3.9), suppose that $pH_2(\alpha )\equiv 0 \pmod {p^2}$ . Then ${\mathcal F}_p(\alpha )\equiv 0\pmod {p^2}$ from (3.11), and hence

(3.12) $$ \begin{align} \alpha^{p}\equiv -\frac{\alpha^{2p}+2}{2} \pmod{p^2}. \end{align} $$

Squaring both sides of (3.12) yields

(3.13) $$ \begin{align} \alpha^{2p}\equiv \frac{(\alpha^{2p}+2)^2}{4} \pmod{p^2}. \end{align} $$

From (3.10 and (3.13),

$$ \begin{align*}-2^{p+2}i^p\equiv 2^{2p}i^{2p}-2^{p+2}i^p+4\pmod{p^2},\end{align*} $$

or equivalently,

(3.14) $$ \begin{align} 2^{2p}i^{2p}+4\equiv 0 \pmod{p^2}. \end{align} $$

Since $i^2\equiv -1 \pmod {p}$ ,

(3.15) $$ \begin{align} i^{2p}\equiv (-1)^p\equiv -1 \pmod{p^2}. \end{align} $$

Consequently, from (3.14) and (3.15),

(3.16) $$ \begin{align} (2^p-2)(2^p+2)\equiv 0 \pmod{p^2}. \end{align} $$

But $2^p\equiv 2 \pmod {p}$ , so $2^p+2\not \equiv 0 \pmod {p}$ since $p\ge 3$ . It follows from (3.16) that $2^p-2\equiv 0 \pmod {p^2}$ and, therefore, $2^{p-1}\equiv 1 \pmod {p^2}$ .

Conversely, if $2^{p-1}\equiv 1 \pmod {p^2}$ , then $2^p-2\equiv 0\pmod {p^2}$ , and the steps of the proof in the other direction can be reversed from (3.16) back to (3.13) to give

(3.17) $$ \begin{align} (\alpha^{2p}+2-2\alpha^p)(\alpha^{2p}+2+2\alpha^p)\equiv 0\pmod{p^2}. \end{align} $$

Suppose that

$$ \begin{align*}\alpha^{2p}+2-2\alpha^p\equiv (\alpha^{2}+2-2\alpha)^p\equiv 0 \pmod{p}.\end{align*} $$

Then, since $H_1(\alpha )=\alpha ^2+2\alpha +2\equiv 0 \pmod {p}$ , we arrive at the contradiction

$$ \begin{align*}H_1(\alpha)-(\alpha^{2}+2-2\alpha)=4\alpha\equiv 0\pmod{p}. \end{align*} $$

Therefore, from (3.17) and (3.11),

$$ \begin{align*}pH_2(\alpha)\equiv {\mathcal F}_p(\alpha) \equiv \alpha^{2p}+2+2\alpha^p \equiv 0 \pmod{p^2},\end{align*} $$

which completes the proof in this direction for the zero $\alpha $ .

To show that we do not need to consider the zero $\beta $ , we now provide a proof of (3.8). Since $\alpha \beta \equiv 2\pmod {p}$ , it follows that

(3.18) $$ \begin{align} \alpha^p\beta^p\equiv 2^p\pmod{p^2}. \end{align} $$

Since $H_1(\beta )\equiv 0 \pmod {p}$ , we see that $\beta ^2\equiv -2(\beta +1) \pmod {p}$ , which implies

(3.19) $$ \begin{align} \beta^{2p}\equiv -2^p(\beta+1)^p\equiv -2^p(p-i)^p\equiv 2^pi^p \pmod{p^2}. \end{align} $$

From the definition of $H_2(x)$ , (2.1) and (3.19), it follows that

(3.20) $$ \begin{align} pH_2(\beta)&=2\beta^p+2-2^p(\beta+1)^p \notag \\&=2\beta^p+2-2^p(p-i)^p \notag \\&\equiv 2\beta^p+2+2^pi^p \pmod{p^2} \notag \\&\equiv 2\beta^p+2+\beta^{2p}\pmod{p^2} \notag \\&\equiv {\mathcal F}_p(\beta) \pmod{p^2}. \end{align} $$

To establish (3.8), suppose that $pH_2(\alpha )\equiv 0 \pmod {p^2}$ . Then ${\mathcal F}_p(\alpha )\equiv 0 \pmod {p^2}$ from (3.11), and therefore, $2^{p}\equiv 2\pmod {p^2}$ from the argument following (3.16). Hence, from (3.18),

$$ \begin{align*} {\mathcal F}_p(\beta)&=\beta^{2p}+2\beta^p+2\\ &\equiv \bigg(\frac{2^p}{\alpha^p}\bigg)^2+2\bigg(\frac{2^p}{\alpha^p}\bigg)+2\pmod{p^2}\\ &\equiv \frac{2(\alpha^{2p}+2\alpha^p+2)}{\alpha^{2p}}\pmod{p^2}\\ &\equiv \frac{2{\mathcal F}_p(\alpha)}{\alpha^{2p}}\pmod{p^2}\\ &\equiv 0 \pmod{p^2}. \end{align*} $$

Thus, $pH_2(\beta )\equiv 0 \pmod {p^2}$ from (3.20). The converse is similar and we omit the details.

4 Comments

The referee suggested a more polynomial-oriented approach, which we outline here. Using the easily verified polynomial congruences

$$ \begin{align*} x^4+4\equiv x^3-2x-4\equiv (x+1)^2+1\equiv 0 \pmod{H_1(x)}, \end{align*} $$

and considering the four cases for $p \pmod {8}$ , we obtain

$$ \begin{align*} pH_2(x)\equiv 2(1+(-1)^{{(p^2+7)}/{8}}2^{{(p-1)}/{2}})((-1)^{{(p^2+4p-5)}/{8}}2^{{(p-1)}/{2}}(x+1) +1) \pmod{H_1(x)}. \end{align*} $$

Since it is straightforward to confirm that no zeros of $(-1)^{{(p^2+4p-5)}/{8}}2^{{(p-1)}/{2}}(x+1) +1$ are zeros of $H_1(x)$ , we conclude that

$$ \begin{align*}pH_2(x)\equiv 0 \pmod{p^2} \quad \text{if and only if} \quad 1+(-1)^{{(p^2+7)}/{8}}2^{{(p-1)}/{2}} \equiv 0 \pmod{p^2},\end{align*} $$

which implies in any case that

$$ \begin{align*}pH_2(x)\equiv 0 \pmod{p^2} \quad \text{if and only if} \quad 2^{p-1}-1 \equiv 0 \pmod{p^2}.\end{align*} $$

The referee asked whether it is possible to generalise Theorem 1.1 to ${\mathcal F}_n(x)$ when n is composite, noting that it seems plausible that ${\mathcal F}_n(x)$ is monogenic if and only if $n$ is not a multiple of a Wieferich prime. We believe that this is the case. However, a proof will require additional ideas because the arguments in our paper use the fact that $p$ is prime in several places.