Prime powers dividing products of consecutive integer values of $x^{2^n}+1$

TL;DR

This paper investigates the prime divisors of products of the form $x^{2^n}+1$ evaluated at consecutive integers, establishing bounds on the prime powers dividing these products and extending previous results for the case $n=1$.

Contribution

It provides new bounds on the orders of primes dividing such products for general $n$, and proves that for $n=2$, these products are never perfect fifth powers or higher.

Findings

Existence of a prime divisor with bounded order for large $m$

For $n=2$, the product is never a fifth power or higher

Extension of Cilleruelo's work from $n=1$ to general $n$

Abstract

Let $n$ be a positive integer and $f(x) := x^{2^n}+1$. In this paper, we study orders of primes dividing products of the form $P_{m,n}:=f(1)f(2)\cdots f(m)$. We prove that if $m > \max\{10^{12},4^{n+1}\}$, then there exists a prime divisor $p$ of $P_{m,n}$ such that ord$_{p}(P_{m,n} )\leq n\cdot 2^{n-1}$. For $n=2$, we establish that for every positive integer $m$, there exists a prime divisor $p$ of $P_{m,2}$ such that ord$_{p} (P_{m,2}) \leq 4$. Consequently, $P_{m,2}$ is never a fifth or higher power. This extends work of Cilleruelo who studied the case $n=1$.

Equations

$$A_f(m):=f(1)f(2)\cdots f(m)$$

$$P_{m,n}:=\prod _{x\le m}\left(x^{2^n}+1\right).$$

$$\pi (x):=\sum _{p\le x}1,\pi (x;q,a):=\sum _{\begin{array}{c}p\le x\\ p\equiv a\mathrm{mod}q\end{array}}1,\text{mbox}and\theta (x;q,a):=\sum _{\begin{array}{c}p\le x\\ p\equiv a\mathrm{mod}q\end{array}}\log p$$

$$\pi (x)\le 1.1\cdot \frac{x}{\log x}.$$

$$\pi (x)\le \left(1+\frac{1.2762}{\log x}\right)\cdot \frac{x}{\log x}\text{mbox}ifx>1.$$

$$\pi (x;2^{n+1},1)\le \frac{4\cdot x}{2^n\log x}.$$

$$\pi (x;q,a)\le \frac{2x}{\phi (q)\log (x/q)}$$

$$\pi (x;2^{n+1},1)\le \frac{2x}{2^n\log (x/2^{n+1})}.$$

$$\begin{gathered} \pi (x;2^{n+1},1)\le \frac{2x}{2^n\log x}\cdot \frac{\log x}{\log (x/2^{n+1})} \\ \le \frac{2x}{2^n\log x}\cdot \frac{\log 4^{n+1}}{\log 2^{n+1}}\le \frac{4x}{2^n\log x} \end{gathered}$$

$$\sum _{\begin{array}{c}p\le x\\ p\equiv a\mathrm{mod}8\end{array}}\frac{\log p}{p}>0.245\log x-\mathrm{3.15.}$$

$$\begin{array}{r}\sum _{\begin{array}{c}p\le x\\ p\equiv a\mathrm{mod}8\end{array}}\frac{\log p}{p}=\frac{\theta (x;8,a)}{x}+\underset{2}{\overset{x}{\int }}\frac{\theta (t;8,a)}{t^2}\text{mbox}dt>\frac{\theta (x;8,a)}{x}+\underset{10^6}{\overset{x}{\int }}\frac{\theta (t;8,a)}{t^2}\text{mbox}dt.\end{array}$$

$$\left|\theta (t;q,a)-\frac{t}{\phi (q)}\right|<0.024\cdot \frac{t}{\log t}\text{mbox}if1\le q\le 10^5,(q,a)=1\text{mbox}andt\ge 10^6.$$

$$\theta (t;8,a)>\frac{t}{4}-0.024\cdot \frac{t}{\log t}$$

$$\sum _{\begin{array}{c}p\le x\\ p\equiv a\mathrm{mod}8\end{array}}\frac{\log p}{p}>\frac{1}{4}-0.024\cdot \frac{1}{\log x}+{\left[\frac{1}{4}\log t-0.024\log \log t\right]}_{10^6}^x.$$

$$\log \log x<0.191\log x.$$

$$N=k_1+k_2+\cdots +k_s(k_1\ge k_2\ge ...\ge k_s)$$

$$\begin{array}{rl}{x}_1^{2^n}+1\equiv & 0\mathrm{mod}{p}^{k_1}\\ ⋮& \\ x_s^{2^n}+1\equiv & 0\mathrm{mod}{p}^{k_s}\end{array}$$

$$p{\ll }_{n}x,$$

$$x:=\max \{|x_1|,...,|x_s|\}.$$

$$x_i^{2^n}+1=\prod _{j=1}^{2^n}\left(x_i+{\alpha }_j\right),$$

$$p|(x_1^{2^n}+1)\Rightarrow \mathfrak{P}|(x_1+{\alpha }_1){\mathcal{O}}_{\mathbb{Q}({\alpha }_1)}$$

$$\tilde{\mathfrak{P}}\nmid (x_1+{\alpha }_1){\mathcal{O}}_{\mathbb{Q}({\alpha }_1)}$$

$$\text{mbox}disc({\mathcal{O}}_{\mathbb{Q}({\alpha }_1)})=\prod _{1\le j_1<j_2\le 2^n}({\alpha }_{j_1}-{\alpha }_{j_2}{)}^2$$

$${\mathfrak{P}}^{k_1}|(x_1+{\alpha }_1){\mathcal{O}}_{\mathbb{Q}({\alpha }_1)}.$$

$${\mathfrak{P}}^{k_i}|(x_i+{\alpha }_{j_i}){\mathcal{O}}_{\mathbb{Q}({\alpha }_1)}$$

$${\mathfrak{P}}^{k_r}|(x_i+{\alpha }_{j_i}){\mathcal{O}}_{\mathbb{Q}({\alpha }_1)}\text{mbox}if1\le i\le r.$$

$${\beta }_1,...,{\beta }_r\in {\mathcal{O}}_{\mathbb{Q}({\zeta }_{2^{m+1}})}$$

$${\beta }_1{\alpha }_{j_1}+\cdots +{\beta }_r{\alpha }_{j_r}=0,$$

$${\beta }_1(x_1+{\alpha }_{j_1})+\cdots +{\beta }_r(x_r+{\alpha }_{j_r})\in {\mathcal{O}}_{\mathbb{Q}({\zeta }_{2^{m+1}})}.$$

$${\mathfrak{P}}^{k_r}|({\beta }_1(x_1+{\alpha }_{j_1})+\cdots +{\beta }_r(x_r+{\alpha }_{j_r})){\mathcal{O}}_{\mathbb{Q}}({\zeta }_{2^{n+1}})$$

Full text

Abstract: Let $n$ be a positive integer and $f(x):=x^{2^{n}}+1$. In this paper, we study orders of primes dividing products of the form $P_{m,n}:=f(1)f(2)\cdots f(m)$. We prove that if $m>\max\{10^{12},4^{n+1}\}$, then there exists a prime divisor $p$ of $P_{m,n}$ such that ${\rm ord}_{p}(P_{m,n})\leq n\cdot 2^{n-1}$. For $n=2$, we establish that for every positive integer $m$, there exists a prime divisor $p$ of $P_{m,2}$ such that ${\rm ord}_{p}(P_{m,2})\leq 4$. Consequently, $P_{m,2}$ is never a fifth or higher power. This extends work of Cilleruelo [6] who studied the case $n=1$.

1. Introduction and main result

For a prime $p$ and a nonzero integer $s$, define ${\rm ord}_{p}(s)$ to be the unique non-negative integer $i$ for which $p^{i}|s$ but $p^{i+1}\nmid s$. Let $f(x)\in\mathbb{Z}[x]$ be a polynomial of degree $k\geq 2$ with positive leading coefficient which does not vanish at any positive integer. Set

[TABLE]

and note that this is nonzero for all $m\in\mathbb{N}$ by the above assumption.

A major unsolved problem in analytic number theory concerns the question whether $f$ represents infinitely many primes if $f$ is irreducible and there exists no prime $p$ dividing $f(m)$ for all integers $m$. If $f$ represents infinitely many primes, then trivially, for infinitely many integers $m$, there exists a prime such that ${\rm ord}_{p}(A_{f}(m))=1$. For particular polynomials $f$, several authors investigated the related question whether for all sufficiently large integers $m$ there exists a prime $p$ with ${\rm ord}_{p}(A_{f}(m))=1$. If this is the case, then, in particular, $A_{f}(m)$ is a perfect power for at most finitely many $m\in\mathbb{N}$.

Below we summarize a number of results from the literature. For the polynomial $f(x)=x^{2}+1$, J. Cilleruelo [6] proved the following result which we shall generalize in this paper.

Theorem 1** (Cilleruelo).**

Let $f(x)=x^{2}+1$ and $m>3$. Then there exists a prime divisor $p$ of $A_{f}(m)$ with ${\rm ord}_{p}(A_{f}(m))=1$. Consequently, $A_{f}(m)$ is a perfect power only for $m=3$, in which case we have $A_{f}(3)=10^{2}$.

Cilleruelo’s work [6] used only elementary tools such as Chebyshev’s upper bound inequality for the primes counting function. In subsequent work by Fang [8], his technique was applied to products $A_{f}(m)$ corresponding to the polynomials $4x^{2}+1$ and $2x^{2}-2x+1$. Yang, Togbé and He [12] proved that for any irreducible quadratic polynomial $f(x)\in\mathbb{Z}[x]$, there exists a prime $p$ with ${\rm ord}_{p}(A_{f}(m))=1$ if $m\geq C$, where $C$ is a computable constant depending on the coefficients of $f(x)$.

Furthermore, the above problem has been investigated by many authors for polynomials of the form $f(x)=x^{k}+1$, where $k$ is an odd positive integer. Gürel and Kisisel [9] settled the case when $k=3$. Based on an idea due to W. Zudilin, Zhang and Wang [13] extended this result to odd primes $k\geq 5$. Recently, Chen et al. [4] managed to handle all odd prime powers $k$. Chen and Gong [3] treated the case when $k$ is a product of at most two odd primes and Dey and Laishram [7] managed to cover all $k$’s which are composed of at most four odd primes.

Thus, for polynomials $f(x)=x^{k}+1$ with $k$ odd, a lot of research has been done. With regard to even $k$’s, the authors are aware only of Cilleruelo’s Theorem, stated above, for the case $k=2$. In this paper, we investigate orders of primes dividing $A_{f}(m)$ for polynomials of the form $f(x)=x^{k}+1$ when $k$ is a power of 2. Note that $f(x)=x^{k}+1$ is irreducible iff $k$ is a power of $2$. Throughout the sequel, we set

[TABLE]

We shall extend Cilleruelo’s Theorem for the case $n=1$ to larger $n$’s as follows.

Theorem 2**.**

Let $n\geq 2$ be an integer. Then there exists a prime divisor $p$ of $P_{m,n}$ with ${\rm ord}_{p}(P_{m,n})\leq n\cdot 2^{n-1}$ if $m>\max\{10^{12},4^{n+1}\}$. Consequently, in this case, $P_{m,n}$ is not a perfect $q$-th power if $q$ is a positive integer exceeding $n\cdot 2^{n-1}$.

For $n=2$, we are able to remove the condition that $m>\max\{10^{12},4^{n+1}\}$, thus obtaining the following.

Theorem 3**.**

For all positive integers $m$, there exists a prime divisor $p$ of $P_{m,2}$ with ${\rm ord}_{p}(P_{m,2})\leq 4$. Consequently, $P_{m,2}$ is never a perfect $q$-th power if $q$ is a positive integer exceeding $4$.

2. Notations and preliminaries

In this section, we provide some inequalities related to primes counting functions which are essential to prove our main results. As usual, we reserve the symbol "$p$" for primes and use the notations

[TABLE]

throughout this paper. Below are the lemmas that we shall use.

Lemma 4**.**

For any $x\geq 10^{6}$, we have

[TABLE]

Proof.

In [5], it was established that

[TABLE]

If $x\geq 10^{6}$, the desired bound follows. ∎

Lemma 5**.**

For any integer $n\geq 2$ and any real $x\geq 4^{n+1}$, we have

[TABLE]

Proof.

The Brun-Titchmarsh inequality, as given by Montgomery and Vaughan [11], asserts that

[TABLE]

whenever $q<x$. This implies

[TABLE]

It follows that

[TABLE]

if $x\geq 4^{n+1}$, which completes the proof. ∎

Lemma 6**.**

For any $x\geq 10^{6}$ and $a\in\{1,3,5,7\}$, we have

[TABLE]

Proof.

Using partial summation, we transform the sum in question into

[TABLE]

By Corollary 1.7. in [2], we have

[TABLE]

This implies

[TABLE]

if $t\geq 10^{6}$. Plugging this into (1), and performing integration, we get

[TABLE]

Since $\frac{\log x}{\log\log x}$ is an increasing function for $x\geq 10^{6}$, we have

[TABLE]

Using this inequality, we obain the desired result from (3). ∎

3. Systems of congruences modulo prime powers

Let $n\in\mathbb{N}$ be given. A significant part of our method consists in finding an as small as possible number $N=N(n)$ such that for every partition

[TABLE]

of $N$ and any distinct $x_{1},...,x_{s}\in\mathbb{N}$ satisfying a system of congruences of the form

[TABLE]

with $p$ an odd prime, it follows that

[TABLE]

where we set

[TABLE]

It will become clear in section 8 how this problem, which is also of independent interest, enters the proof of Theorems 2 and 3. The following sections 4 to 7 are dedicated to solving this problem. The result will lead us directly to the quantity $n\cdot 2^{n-1}$ in Theorem 2.

4. Reformulation in Cyclotomic fields

For a number field $K$, we denote by $\mathcal{O}_{K}$ the ring of algebraic integers in $K$. If $L$ is a finite extension of the number field $K$ and $a\in L$, we denote by $N_{L:K}(a)$ the norm of $a$ over $K$.

We write

[TABLE]

where $\left\{\alpha_{1},...,\alpha_{2^{n}}\right\}$ is the set of primitive $2^{n+1}$-th roots of unity. Then

[TABLE]

for some prime ideal $\mathfrak{P}$ in $\mathcal{O}_{\mathbb{Q}(\alpha_{1})}$ lying over $p$, but

[TABLE]

for any prime ideal $\tilde{\mathfrak{P}}\not=\mathfrak{P}$ conjugate to $\mathfrak{P}$. Otherwise, $\tilde{\mathfrak{P}}$ would divide both the ideals $(x_{1}+\alpha_{1})\mathcal{O}_{\mathbb{Q}(\alpha_{1})}$ and $(x_{1}+\alpha_{j})\mathcal{O}_{\mathbb{Q}(\alpha_{1})}$ for some $j\not=1$ and hence the ideal $(\alpha_{1}-\alpha_{j})\mathcal{O}_{\mathbb{Q}(\alpha_{1})}$. However, this is not possible because $p>2$ and the discriminant

[TABLE]

of $\mathcal{O}_{\mathbb{Q}(\alpha_{1})}$ has no rational prime divisors other than 2. Hence, from the first congruence in (5), it follows that

[TABLE]

If $i\in\{2,...,s\}$, then $\mathfrak{P}|(x_{i}+\alpha_{j_{i}})\mathcal{O}_{\mathbb{Q}(\alpha_{1})}$ for some unique $j_{i}\in\{1,...,2^{n}\}$, and we have

[TABLE]

from the $i-th$ congruence in (5) by a similar argument as above. Set $j_{1}:=1$. Since $k_{1}\geq k_{2}\geq...\geq k_{s}$, it follows that for every $r\in\{1,...,s\}$, we have

[TABLE]

Now let $m<n$ be a non-negative integer and $r\in\{1,...,s\}$. Denote by $\zeta_{k}$ a primive $k$-th root of unity. If

[TABLE]

are such that

[TABLE]

then

[TABLE]

Since also

[TABLE]

using (6), it follows that

[TABLE]

and hence

[TABLE]

If in addition

[TABLE]

then we deduce that

[TABLE]

and hence

[TABLE]

which implies

[TABLE]

provided that

[TABLE]

Clearly, we also have the bound

[TABLE]

if

[TABLE]

Now we consider an arbitrary non-negative integer $m$. If $m\geq n$, then we set $R(m,n):=1$. If $m<n$, then let $R(m,n)$ be the smallest number $r$ such that given any $r$ primitive $2^{n+1}$-th roots of unity $\gamma_{1},...,\gamma_{r}$ (not necessarily distinct), there exist

[TABLE]

such that

[TABLE]

and

[TABLE]

for any distinct positive integers $x_{1},...,x_{r}$. Then it follows that

[TABLE]

provided that for any partition of the form in (4) we have

[TABLE]

for some $r\in\{1,...,s\}$ and $m\in\{1,...,n\}$.

Note that $R(m,n)$ decreases as $m$ increases. Hence, we may choose

[TABLE]

and our above condition reduces to

[TABLE]

5. Determining $R(m,n)$

Throughout the sequel, for any real number $x$, we denote by $\lceil x\rceil$ the smallest integer greater or equal $x$ and by $\lfloor x\rfloor$ the largest integer less or equal $x$. We now prove the following.

Lemma 7**.**

For any natural numbers $m$ and $n$, we have

[TABLE]

Proof.

This is trivial if $m\geq n$. So assume $m<n$. Then we claim that among $2^{n-m-1}+1$ (not necessarily distinct) primitive $2^{n+1}$-th roots of unity, there exist two, $\gamma_{1}$ and $\gamma_{2}$, such that $\beta:=\gamma_{1}/\gamma_{2}$ is a $2^{m+1}$-th root of unity. This is equivalent to saying that among $2^{n-m-1}+1$ (not necessarily distinct) odd integers in $\{1,3,...,2^{n+1}-1\}$, there exist two whose difference is divisible by $2^{n-m}$. Indeed, these integers fall into $2^{n-m-1}$ possible residue classes modulo $2^{n-m}$. By pigeonhole principle, two of them fall into the same residue class, and hence the claim follows.

Now, for $\gamma_{1}$, $\gamma_{2}$, $\beta$ as above, we have $\gamma_{1}-\beta\gamma_{2}=0$, but clearly $x_{1}-\beta x_{2}\not=0$ for any two distinct positive integers $x_{1}$ and $x_{2}$. This proves that

[TABLE]

It remains to show that

[TABLE]

Assume the contrary. We look at the example

[TABLE]

These are primive $2^{n+1}$-th roots of unity. Moreover, the set

[TABLE]

forms an integral basis of $\mathbb{Q}(\zeta_{2^{n}})$ over $\mathbb{Q}(\zeta_{2^{m+1}})$. Hence, if $\beta_{1},...,\beta_{2^{n-m-1}}\in\mathcal{O}_{\mathbb{Q}(\zeta_{2^{m+1}})}$, then

[TABLE]

implies $\beta_{1}=...=\beta_{2^{n-m-1}}=0$. But then our second condition

[TABLE]

is violated for any integers $x_{1},...,x_{2^{n-m-1}}$. This gives a contradiction. Hence, (12) follows which completes the proof. ∎

Remark. For our purposes, it would have been sufficient to prove that $R(m,n)\leq 2^{n-m-1}+1$.

6. Transformation into a combinatorial condition

By the considerations in the previous section, an admissible $N(n)$ is the smallest integer $N$ such that every partition of the form in (4) satisfies the condition

[TABLE]

In the following, we construct an extreme partition satisfying (13). We take $k_{1},...,k_{s}$ as large as possible such that (13) is not satisfied if $1\leq r\leq s-1$ but satisfied if $r=s$. This property determines the partition in question completely, namely we obtain

[TABLE]

[TABLE]

and

[TABLE]

Moreover, we calculate that

[TABLE]

Examples: For $n=1,2,3,4,5$, we get the partitions

[TABLE]

In the following, we prove that this actually gives exactly the minimal number $N$ we are aiming for.

Lemma 8**.**

The number

[TABLE]

is the smallest positive integer $N$ such that every partition of the form in (4) satisfies the condition (13).

Proof.

Look at the extreme partition constructed above. $N$ cannot be chosen smaller because

[TABLE]

is a partition of $n\cdot 2^{n-1}$ which does not satisfy the required condition. Now, if

[TABLE]

is any partition different from our extreme partition above, then $s^{\prime}\geq s$ or $k_{r}^{\prime}>k_{r}$ for some $r\in\{1,...,s^{\prime}\}$. In the first case,

[TABLE]

In the second case,

[TABLE]

by construction of the partition $N=k_{1}+...+k_{s}$. Hence, our new partition $N=k_{1}^{\prime}+...+k_{s^{\prime}}^{\prime}$ satisfies the desired condition. This completes the proof. ∎

7. Back to congruences modulo prime powers

Now we are ready to prove what we formulated as a goal in section 3. In addition, we observe that we even get an upper bound for $p$ which does only depend on $x$ and not on $n$. Indeed, taking our proof of Lemma 7 in consideration, we may choose $\beta_{u}=1$ and $\beta_{v}=-\beta$ with $\beta$ a $2^{m+1}$-th root of unity for suitable distinct $u,v\in\{1,...,r\}$ and $\beta_{i}=0$ if $i\not\in\{1,...,r\}\setminus\{u,v\}$. This implies

[TABLE]

for any $\sigma\in$ Gal$(\mathbb{Q}(\zeta_{2^{n+1}}):\mathbb{Q})$ and hence

[TABLE]

Upon recalling that $\beta_{1}(x_{1}+\alpha_{j_{1}})+\cdots+\beta_{r}(x_{r}+\alpha_{j_{r}})\in\mathcal{O}_{\mathbb{Q}(\zeta_{2^{m+1}})}$, it follows that

[TABLE]

Under the condition (8) which says that $k_{r}\geq 2^{m}$, it follows now from (7) that

[TABLE]

Similarly, we find that

[TABLE]

under the condition (9) which says that $k_{r}\geq 2^{n}$.

Summarizing our results in the previous sections and taking our observation above into account, we thus have established the following.

Theorem 9**.**

Let $n\in\mathbb{N}$ be given. Set $N=N(n):=n\cdot 2^{n-1}+1$. Then if

[TABLE]

is any partition of $N$ and the system of congruences

[TABLE]

holds for distinct $x_{1},...,x_{s}\in\mathbb{N}$ and $p$ an odd prime, then

[TABLE]

where $x:=\{|x_{1}|,...,|x_{s}|\}$.

8. proof of Theorem 2

Let us assume that $n\geq 2$, $m>\max\{10^{12},4^{n+1}\}$ and

[TABLE]

for all primes $p$ dividing $P_{m,n}$. Then from Theorem 9, it follows that

[TABLE]

for all these primes. (This will be essential in our proof.) Hence, we can write $P_{m,n}$ as

[TABLE]

where the ${\alpha}_{p}$’s are non-negative integers with either ${\alpha}_{p}=0$ or ${\alpha}_{p}>n\cdot 2^{n-1}$. Clearly,

[TABLE]

Write

[TABLE]

where the ${\beta}_{p}$’s are positive integers.

Combining (16), (17) and (18), we have

[TABLE]

Taking logarithm, if follows that

[TABLE]

Since

[TABLE]

we see that

[TABLE]

Now let $p$ be an odd prime dividing $x^{2^{n}}+1$. Then $p\equiv 1\bmod{2^{n+1}}$ and, moreover, there are exactly $2^{n}$ solutions to the congruence

[TABLE]

(Note that the $2^{n+1}$-th cyclotomic field $\mathbb{Q}(\zeta_{2^{n+1}})$ is the splitting field of the polynomial $x^{2^{n}}+1$ over $\mathbb{Q}$, and the rational primes which split completely in this field are exactly those congruent to $1\bmod{2^{n+1}}$.) By Hensel’s lemma, they extend uniquely to solutions of

[TABLE]

for any $j\geq 1$. Thus, each interval of length $p^{j}$ contains exactly $2^{n}$ solutions of this congruence. It follows that

[TABLE]

Also we have

[TABLE]

From (21) and (22), we deduce that

[TABLE]

Combining inequalities (19), (20) and (23), and recalling that $\alpha_{p}=0$ if $p>2$ and $p\not\equiv 1\bmod{2^{n+1}}$, we obtain

[TABLE]

If $p>m$, then from (21), we have

[TABLE]

since

[TABLE]

Moreover, from (22), we have

[TABLE]

where $M=\lfloor\frac{\log m}{\log p}\rfloor$. Hence, for $p\leq m$, we deduce that

[TABLE]

Combining inequalities (24), (25) and (26), we get

[TABLE]

which implies

[TABLE]

Now recalling that $n\geq 2$ and using Lemma 6, we have

[TABLE]

Combining inequalities (27) and (28), applying Lemmas 5, and dividing by $m-1$, we obtain

[TABLE]

Note that the limit, as $m\rightarrow\infty$, of the right-hand side is

[TABLE]

Hence, if $m$ is large enough, then the above inequality will be false. An easy calculation shows that this is the case whenever $m>10^{12}$ and hence we reach a contradiction. We conclude that there exists a prime $p$ with ${\rm ord}_{p}(P_{m,n})\leq n\cdot 2^{n-1}$, which completes the proof.

Remark: We note that the above argument would not go through if we had some much weaker condition like $p<\beta n^{\alpha}$ with $\alpha>1$ in place of (15).

9. Proof of Theorem 3

By Theorem 2, there exists a prime $p$ with ${\rm ord}_{p}P_{m,2}\leq 4$ if $m>10^{12}$. Therefore, it suffices to check that the same holds if $1\leq m\leq 10^{12}$. The claim is trivial if $1\leq m\leq 5$ since $1\leq{\rm ord}_{2}(P_{m,2})\leq 3$. Further, we observe that $6^{4}+1=1297$ is a prime and the next $x$’s for which $1297$ divides $x^{4}+1$ are $x=216,1081,1291,1303$. Moreover, for these $x$’s, ${\rm ord}_{1297}(x^{4}+1)=1$. Hence, there exists a prime $p$ with ${\rm ord}_{p}(P_{m,2})\leq 4$ if $6\leq m\leq 1302$. Next, we observe that $1302^{4}+1=2873716601617$ is a prime as well. The next $x$’s for which $2873716601617$ divides $x^{4}+1$ are $x=2207155608,2871509446009,2873716600315,2873716602919$. Moreover, for these $x$’s, ${\rm ord}_{2873716601617}(x^{4}+1)=1$. Hence, we conclude that there exists a prime $p$ with ${\rm ord}_{p}(P_{m,2})\leq 4$ if $1302\leq m\leq 2873716602918$. This completes the proof.