Optimal bounds for B\"uchi's problem in modular arithmetic II

TL;DR

This paper establishes upper bounds on the length of sequences of squares generated by quadratic polynomials over modular rings, extending B"uchi's problem to higher powers of primes.

Contribution

It provides the first bounds for B"uchi's problem in modular arithmetic for prime powers, generalizing previous results for prime moduli.

Findings

Existence of a uniform bound M for sequences of squares from quadratic polynomials over Z/p^sZ.

Reduction to polynomials with invertible dominant coefficients simplifies the analysis.

The bounds depend only on p, s, and the polynomial's properties.

Abstract

Given a prime $p\ge5$ and an integer $s\ge1$, we show that there exists an integer $M$ such that for any quadratic polynomial $f$ with coefficients in the ring of integers modulo $p^s$, such that $f$ is not a square, if a sequence $(f(1),\dots,f(N))$ is a sequence of squares, then $N$ is at most $M$. We obtain this result by reducing to the case where $f$ has an invertible dominant coefficient.

Equations

$$B_f^N=(f(1),\dots ,f(N))$$

$$p^{\frac{s-t_2}{2}}-1,$$

$$\underset{f_0}{\max }\left\{N:{\left[B_f^N\right]}_m\text{ is a non-trivial Bu¨chi sequence, where }f=f_2{X}^2+f_{1}X+f_0\right\},$$

$$\mathrm{opt}(m,f_2,f_1)=\mathrm{ml}(m,f_2,f_1)+1.$$

$$\mathrm{ml}(m,f_2)=\underset{f_1}{\max }\mathrm{ml}(m,f_2,f_1)\text{and}\mathrm{opt}(m,f_2)=\mathrm{ml}(m,f_2)+1.$$

$$\mathrm{ml}(p,0)=\underset{f_1}{\max }\mathrm{ml}(p,0,f_1)=\max \{\mathrm{ml}(p,0,f_1):[f_1{]}_p\ne [0{]}_p\}.$$

$$\mathrm{opt}(p^s,f_2)=\{\begin{array}{ll}\mathrm{opt}(p,0)& \text{ if t2 is odd or t2=∞,}\\ \max \{\mathrm{opt}(p,0),\mathrm{opt}(p^{s-t_2},g_2)\}& \text{ if t2 is even.}\end{array}$$

$$\underset{a,\nu }{\max }\left\{N:{\left[B_f^N\right]}_m\text{ is an f-Bu¨chi sequence, where }f=\alpha (X+\nu {)}^2+a\text{ and }a\ne 0\right\},$$

$${\mathrm{ord}}_p{f}_0<\min \{{\mathrm{ord}}_p{f}_1,{\mathrm{ord}}_p{f}_2\}$$

$$f\equiv (p^r{g}_0{)}^2+p^{2r+1}Xg\equiv (p^r{g}_0{)}^2(1+pXh)(\mathrm{mod}{p}^s)$$

$$2u<s,$$

$$g=g_0+g_{1}X+p^{v}h{X}^2,$$

$$\phi =p^u(g_0+g_{1}X+p^{v}h{X}^2).$$

$$2u={\mathrm{ord}}_p{f}_0<\min \{{\mathrm{ord}}_p{f}_1,{\mathrm{ord}}_p{f}_2\}.$$

$$f\equiv p^{2u}({\ell }^2+p^{v}h{X}^2(2\ell +p^{v}h{X}^2))(\mathrm{mod}{p}^s),$$

$$f-p^{2u}{\ell }^2\equiv p^{2u+v}h{X}^2(2\ell +p^{v}h{X}^2)(\mathrm{mod}{p}^s).$$

$$2u+v<s,$$

$$(2\ell {)}^{s-1}-(2\ell {)}^{s-2}{p}^{v}h{X}^2+\cdots +(-1{)}^{s-1}(p^{v}h{X}^2{)}^{s-1}$$

$$(f-p^{2u}{\ell }^2)[(2\ell {)}^{s-1}-(2\ell {)}^{s-2}{p}^{v}h{X}^2+\dots ]$$

$$\equiv p^{2u+v}h{X}^2(2\ell {)}^s(\mathrm{mod}{p}^s),$$

$$\left(\tilde{f}-{\ell }^2\right)[(2\ell {)}^{s-1}-(2\ell {)}^{s-2}{p}^{v}h{X}^2+\cdots ]\equiv p^{v}h{X}^2(2\ell {)}^s(\mathrm{mod}{p}^{v+1}),$$

$$\left(\tilde{f}-{\ell }^2\right)(2\ell {)}^{s-2}{p}^{v}h{X}^2$$

$$2p^v{h}_0{g}_0\cdot 2^{s-2}{g}_1^{s-2}{p}^v{h}_d,$$

$$p^v{h}_d{2}^s{g}_1^s,$$

$$p^v\cdot 2h_0{g}_0\cdot 2^{s-2}{g}_1^{s-2}{h}_d\equiv h_d{2}^s{g}_1^s(\mathrm{mod}p),$$

$$\min \{{\mathrm{ord}}_p{f}_1,{\mathrm{ord}}_p{f}_2\}>{\mathrm{ord}}_p{f}_0$$

$$x^2+k{p}^s=f(1)=p^{t_0}g(1)\equiv p^{t_0}{g}_0=f_0(\mathrm{mod}{p}^{t_0+1}),$$

$$\mathrm{opt}(p^s,f_2,f_1)=\mathrm{opt}(p,0,f_1)$$

$$f=f_2{X}^2+f_{1}X+b\equiv f_{1}X+b(\mathrm{mod}p).$$

$$f=f_2{X}^2+f_{1}X-f_1{x}_0-f_2{x}_0^2\equiv f_{1}X+b(\mathrm{mod}p),$$

$$\mathrm{opt}(p^s,f_2,f_1)=\mathrm{opt}(p^{s-m},p^{t_2-m}{g}_2,p^{t_1-m}{g}_1)$$

Full text

Keywords: Büchi sequence, Hensley sequence

MSC 2010 Classification: 11B50, 11B83

1 Introduction

We are interested in the following question:

Question 1.1**.**

Given an integer $m\geq 3$ and a quadratic polynomial $f(x)=f_{2}x^{2}+f_{1}x+f_{0}$ over $\mathbb{Z}$, consider the sequence

[TABLE]

modulo $m$. How long can this sequence be if every element of it is a square modulo $m$ but $f$ itself is not a square modulo $m$?

The same question can be considered for any commutative ring $R$ with unit instead of a quotient of $\mathbb{Z}$. For $R=\mathbb{Z}$, it was first asked by R. Büchi in the early seventies, and was motivated by a decision problem in Logic — see [Lip90] and [Maz94].

In the case of modular arithmetic, Question 1.1 was first addressed by D. Hensley in [He, Thm. 3.1]. He showed that in the particular case where $m$ is an odd prime number and $f(x)$ is of the form $(x-\nu)^{2}-a$, then $N$ is strictly less than $m$ — nevertheless he does not give any explicit formula for the largest possible $N$ as $\nu$ and $a$ vary. In [SVV15], we deal with the case where, for an odd given $m$, $f_{2}$ is invertible modulo $m$ — see Theorem 2.3 below for the case of prime powers, and [SVV15, Section 5] for general $m$. In the present paper, we solve the problem for any given prime power and any $f$, by reducing it to the case where $f_{2}$ is invertible — see Theorem 2.2 below.

To give a taste of our main result without introducing too many technicalities, here we state a corollary.

Theorem 1.2**.**

Let $p$ be a prime $\geq 3$. Assume that $g_{2}$ is a non-zero square modulo $p$, $s$ and $t_{2}<s$ are positive even integers, and $f_{2}=p^{t_{2}}g_{2}$. As $f$ varies within the set of non-square quadratic polynomials with this restriction on $f_{2}$, the largest possible $N$ such that each of $f(1)$, …, $f(N)$ is a square modulo $p^{s}$, is

[TABLE]

(i.e.: there are sequences of this length, and no larger ones).

Let us give a concrete example. For any odd prime $p$, modulo $p^{4}$, the polynomial $f(x)=p^{2}x^{2}+p^{3}$ is not the square of a polynomial, because $p^{3}$ is not a square, but by Hensel’s lemma, it is easy to see that $f(k)$ is a square modulo $p^{4}$ for $k=1,\dots,p-1$, so the length $p^{\frac{4-2}{2}}-1=p-1$ is reached. Note that $f(0)$ and $f(p)$ are not squares modulo $p^{4}$.

Though many analogous results exist in the literature over different type of rings $R$, they always assume that the polynomial $f$ is monic, with just one exception: Natalia Garcia-Fritz [Ga17, Thm. 1.6, Cor. 1.7 and the comments that follow] does not put restrictions on $f_{2}$ (unconditionally if $R$ is a function field of a curve over $\mathbb{C}$, and assuming the Bombieri-Lang conjecture when $R=\mathbb{Q}$). There is some literature on sequences of squares whose second difference is an arbitrary element of $R$, which corresponds essentially to considering a quadratic $f$ with an arbitrary dominant coefficient. Symmetric sequences of that kind were considered by Allison [All86], Bremner [Bre03], Browkin and Brzeziński [BB06], and Gonzalez-Jimenez and Xarles [GoX11].

Analogues of Question 1.1 have been considered for most classical rings (but in the case of number fields, under some well-known conjectures, like Bombieri-Lang for surfaces, or some version of ABC). Relevant results in positive characteristic can be found in [Pa11] (the analogue of Büchi’s problem for any power over fields with a prime number of elements), and in [PaW15] (over rings of functions), who generalize previous results in [PhV06, PhV10, ShV10, AW11, AHW13]. For a general survey on Büchi’s problem and its extensions to other structures and higher powers, see [PaPhV10].

2 Preliminaries and Main result

If $n$ is an integer, $[n]_{m}$ will denote its residue class modulo $m$ (we may use the brackets notation for polynomials and for sequences as well), and if $p$ is a prime, ${\rm ord}_{p}n$ will stand for the usual order at $p$ of $n$, with the convention ${\rm ord}_{p}(0)=\infty$, so that for every integer $x$ we have, ${\rm ord}_{p}x<\infty$ if and only if $x\neq 0$.

From now on, we will only consider sequences ${B}_{f}^{N}$ over $\mathbb{Z}$ which satisfy the two following conditions for some odd integer $m\geq 3$:

1. (C1)

$f(1)$, …, $f(N)$ are squares modulo $m$. 2. (C2)

$f$ is not the square of a polynomial modulo $m$.

Following [SVV15], sequences ${B}_{f}^{N}$ satisfying (C1) are called $f$-Büchi sequences modulo $m$, and they are called non-trivial if (C2) is also satisfied (we may say just “Büchi” instead of “$f$-Büchi” when it is clear what $f$ is). We should immediately point out that in [SVV15] we consider $f$-Büchi sequences as trivial when $f$ is the square of a polynomial of degree at most $1$. But indeed, when $f_{2}$ is invertible this makes no difference with condition (C2), as shown by the following proposition, which will be proved at the beginning of the next section.

Proposition 2.1**.**

Let $p$ be an odd prime and $s$ be a positive integer. Let $f=f_{2}X^{2}+f_{1}X+f_{0}\in\mathbb{Z}[X]$. The following statements are equivalent:

The polynomial $f$ is the square of a polynomial modulo $p^{s}$. 2. 2.

Either ${\rm ord}_{p}f_{0}<\min\{{\rm ord}_{p}f_{1},{\rm ord}_{p}f_{2}\}$ and $[f_{0}]_{p^{s}}$ is a square, or $f$ is the square of a polynomial modulo $p^{s}$ whose degree is at most one.

For odd $m\geq 3$, let us write ${\rm ml}(m,f_{2},f_{1})$ for

[TABLE]

(with the convention ${\rm ml}(m,f_{2},f_{1})=0$ if all the sequences in the set are trivial), and

[TABLE]

Also, we will write

[TABLE]

Note that when $m=p$ is prime, we trivially have

[TABLE]

Here “${\rm ml}$” stands for “maximal length” and “${\rm opt}$” stands for “optimal bound”. The reason to use both concepts is that the proofs are done in terms of maximal lengths but the formulas that we need from [SVV15] are nicer in terms of the optimal bound (the reader will see the point in Theorem 2.3 below).

We can now state our main theorem.

Theorem 2.2**.**

Let $p$ be a prime $\geq 3$ and $s$ be a positive integer. Let $f_{2}\in\mathbb{Z}$, with $f_{2}\notin p^{s}\mathbb{Z}$ unless $f_{2}=0$. Write $t_{2}={\rm ord}_{p}f_{2}$ and let $g_{2}$ be such that $f_{2}=p^{t_{2}}g_{2}$ when $f_{2}\neq 0$, and $g_{2}=0$ otherwise. Assume $t_{2}\neq 0$. We have

[TABLE]

In this paper, we deal only with prime power modulus. The case of a general modulus $m$ can be reduced to the case of powers of primes following the strategy described in [SVV15, Section 5.1]. Any Büchi sequence modulo $m=p_{1}^{s_{1}}\dots p_{k}^{s_{k}}$ can be glued from Büchi sequences modulo the $p_{i}^{s_{i}}$ using the Chinese Remainder Theorem. The only subtle point to take care of is that one has to check that the resulting sequence modulo $m$ is non trivial, so there are various cases to consider, which result in an elementary but cumbersome analysis — we leave the details to the reader.

While Theorem 2.2 is a natural extension of what we did in our previous work, it was not clear from the beginning what the right statement should be (for example, we were surprised when we discovered that the concept of triviality had to be kept unchanged). We have tried to write the proof in the most uniform possible way, instead of doing the obvious case by case analysis.

In order to have a global picture of the situation, and for later references, we resume in a single theorem what we knew in the case where $f_{2}$ is invertible. In [SVV15], given $\alpha\in\mathbb{Z}$ not divisible by $p$, we had defined ${\rm ml}(m,\alpha)$ as

[TABLE]

and ${\rm opt}(m,\alpha)={\rm ml}(m,\alpha)+1$. Note that when $f_{2}$ is invertible modulo an odd $m\geq 3$, then every polynomial $f(X)=f_{2}X^{2}+f_{1}X+f_{0}$ can be written in a unique way in the form $\alpha(X+\nu)^{2}+a$ modulo $m$, so the notation in [SVV15] is compatible with the present one.

Theorem 2.3**.**

([SVV15, Thm. 1.7, Thm. 1.8, Lem. 2.13])* Let $p$ be an odd prime number, and $s\geq 1$ and $f_{2}$ be integers. Assume that $[f_{2}]_{p^{s}}$ is invertible.*

If $[f_{2}]_{p^{s}}$ is a non-square and $p\geq 5$, then ${\rm opt}(p^{s},f_{2})={\rm opt}(p,n)<\infty$, where $n$ is any quadratic non-residue modulo $p$. 2. 2.

If $[f_{2}]_{p^{s}}$ is a non-zero square and $s=2r$ is even, then ${\rm opt}(p^{s},f_{2})=p^{r}$. 3. 3.

If $[f_{2}]_{p^{s}}$ is a non-zero square and $s=2r+1$ is odd, then ${\rm opt}(p^{s},f_{2})={\rm opt}(p,1)p^{r}<\infty$. 4. 4.

We have ${\rm opt}(p,0)\leq\frac{p+3}{2}$. 5. 5.

For any $k\in\mathbb{Z}$, we have ${\rm opt}(3,2+3k)=\infty$. 6. 6.

For any $s\geq 2$ and $k\in\mathbb{Z}$, we have ${\rm opt}(3^{s},2+3k)=5$.

We get Theorem 1.2 by first applying Theorem 2.2 and then Theorem 2.3, items 2 and 4. For other cases, it is clear how similar corollaries can be obtained.

3 Reduction to the case when $[f_{2}]_{p^{s}}$ is invertible or is $[0]_{p^{s}}$

We will frequently use the following well known fact.

Lemma 3.1**.**

Let $p$ be an odd prime number. If $y\in\mathbb{Z}$ is a non-zero square modulo $p^{t}$ for some $t\geq 1$, then $y$ is a square modulo $p^{s}$ for any $s\geq 1$.

Next, we prove Proposition 2.1.

Proof of Proposition 2.1.

We first prove that 2 implies 1. Assume

[TABLE]

and $[f_{0}]_{p^{s}}$ is a square. If $f$ is identically [math] modulo $p^{s}$, then the claim is trivial, so we may assume that $f_{0}$ is not [math] modulo $p^{s}$. We have

[TABLE]

for some $g\in\mathbb{Z}[X]$, $g_{0}\in\mathbb{Z}$ not divisible by $p$, and $h\in\mathbb{Z}[X]$ such that $g_{0}^{2}h\equiv g\pmod{p^{s}}$. The Taylor series modulo $p^{s}$ of the square root of $1+pXh$ is actually a polynomial, since denominators are powers of $2$ and numerators have increasing order at $p$.

We now prove that 1 implies 2. Assume $s\geq 2$ (indeed, for $s=1$, the claim is trivial as $\mathbb{F}_{p}$ is an integral domain). Let $\varphi\in\mathbb{Z}[X]$ be such that $[f]_{p^{s}}=[\varphi^{2}]_{p^{s}}$. We can assume $[\varphi]_{p^{s}}\neq[0]_{p^{s}}$. Let $u$ be the largest integer such that $\varphi=p^{u}g$ and $g\in\mathbb{Z}[X]$, so that $[g]_{p}\neq[0]_{p}$. We have $f\equiv\varphi^{2}\equiv p^{2u}g^{2}\pmod{p^{s}}$. If $2u\geq s$, there is nothing to prove, so we can assume

[TABLE]

hence $2u+1\leq s$. Let $\tilde{f}=\tilde{f}_{0}+\tilde{f}_{1}X+\tilde{f}_{2}X^{2}\in\mathbb{Z}[X]$ be such that $p^{2u}\tilde{f}=f$ and $[g^{2}]_{p}=[\tilde{f}]_{p}$. So $[g^{2}]_{p}$ has degree at most $2$, and since $[g]_{p}\neq[0]_{p}$, we deduce that $[g]_{p}$ has degree at most $1$ (because we are now over the integral domain $\mathbb{F}_{p}$). So we have

[TABLE]

for some $v\geq 1$ and some $h\in\mathbb{Z}[X]$. If $h$ is the zero polynomial, then we are done. Otherwise choose $v$ as large as possible, so that $h$ has at least one coefficient not divisible by $p$, namely, $[h]_{p}\neq[0]_{p}$. We have then

[TABLE]

Case 1: Assume that $p$ divides $g_{1}$, so that $p$ does not divide $g_{0}$. We have then $\varphi=p^{u}g_{0}+p^{u+1}k_{0}$ for some $k_{0}\in\mathbb{Z}[X]$, hence $f\equiv\varphi^{2}\equiv p^{2u}g_{0}^{2}+p^{2u+1}k_{1}\pmod{p^{s}}$ for some $k_{1}\in\mathbb{Z}[X]$, so

[TABLE]

Case 2: Assume that $p$ does not divide $g_{1}$. Write $\ell=g_{0}+g_{1}X$, so that

[TABLE]

hence

[TABLE]

If $2u+v\geq s$, then we are done (since $\ell$ has degree $1$), so it remains to consider the case where

[TABLE]

which will turn out to be impossible. Multiplying both sides of Equation (1) by

[TABLE]

we obtain:

[TABLE]

since $v\geq 1$. Hence we have

[TABLE]

since $s-2u\geq v+1$. We now compare the coefficients of $X^{2+d+s}$ on both sides, where $d$ is the degree of $[h]_{p}$. Let $h_{d}\in\mathbb{Z}$ be the coefficient of $h$ at $X^{d}$ (so $[h_{d}]_{p}$ is the dominant coefficient of $[h]_{p}$), and let $h_{0}$ be the constant term of $h$. The coefficient of $X^{2+d+s}$ modulo $p^{v+1}$ on the left hand side is the coefficient of

[TABLE]

modulo $p^{v+1}$ (the terms that are not written in Equation (2) will have order at least $2v\geq v+1$, and in the term $(\tilde{f}-\ell^{2})(2\ell)^{s-1}$ all monomials have degree at most $s+1<2+d+s$), which is

[TABLE]

(here the term $2p^{v}h_{0}g_{0}$ comes from Equation (1)). On the other hand, the right hand side of Equation (2) gives

[TABLE]

so we have

[TABLE]

which is a contradiction since $p$ divides neither $g_{1}$ nor $h_{d}$. ∎

Note that if any of $f_{1}$ or $f_{2}$ is invertible modulo $p$, then the condition

[TABLE]

is never satisfied. We now prove a sequence of lemmas that will imply our main theorem.

Lemma 3.2**.**

Let $p$ be an odd prime and $s\geq 1$. Let $f(X)=f_{2}X^{2}+f_{1}X+f_{0}\in\mathbb{Z}[X]$. Write $t_{i}={\rm ord}_{p}f_{i}$ for each $i$. Assume that $\min\{t_{1},t_{2}\}>t_{0}$ and $[f_{0}]_{p^{s}}$ is a non-square. If ${B}_{f}^{N}$ is a Büchi sequence modulo $p^{s}$, then $N=0$.

Proof.

Note that $t_{0}\neq\infty$. Also note that $t_{0}<s$ (because $[f_{0}]_{p^{s}}$ is a non-square, so in particular it is not $[0]_{p^{s}}$). Write $f(X)=p^{t_{0}}g(X)$, where $g(X)=p^{t_{2}-t_{0}}g_{2}X^{2}+p^{t_{1}-t_{0}}g_{1}X+g_{0}$, so that $g_{0}$ is invertible modulo $p$. We assume $N\geq 1$ and will get a contradiction. Let $k,x\in\mathbb{Z}$ be such that $f(1)=x^{2}+kp^{s}$. From the hypothesis of the lemma, we have

[TABLE]

hence, recalling that $t_{0}+1\leq s$, $f_{0}$ is a square modulo $p^{t_{0}+1}$. Since $g_{0}$ is non-zero modulo $p$, also $f_{0}$ is non-zero modulo $p^{t_{0}+1}$. So $f_{0}$ is a square modulo $p^{s}$ by Lemma 3.1, which contradicts our hypothesis on $f_{0}$. ∎

Lemma 3.3**.**

Let $p$ be an odd prime and $s\geq 1$. Let $f_{1},f_{2}\in\mathbb{Z}$. If $f_{1}$ is invertible modulo $p$ and $f_{2}$ is not invertible modulo $p$, then

[TABLE]

Proof.

We first prove the “$\leq$” inequality. Let $N\geq 0$ and $f_{0}$ be integers. Write $f=f_{2}X^{2}+f_{1}X+f_{0}$ and assume that ${B}_{f}^{N}$ is a non-trivial Büchi sequence modulo $p^{s}$. Modulo $p$, since $f_{2}$ is not invertible, we have $f\equiv f_{1}X+f_{0}$. Write $g=f_{1}X+f_{0}$. Since $f(x)$ is a square modulo $p^{s}$ for each $x=1,\dots,N$, it is a square modulo $p$, so ${B}_{g}^{N}$ is a Büchi sequence modulo $p$. Since $f_{1}$ is invertible, ${B}_{g}^{N}$ is a non-trivial Büchi sequence modulo $p$, hence $N$ is at most ${\rm ml}(p,0,f_{1})$.

We now prove the other inequality. Let $h=f_{1}X+b$ be such that ${B}_{h}^{N}$ is a Büchi sequence of length $N={\rm ml}(p,0,f_{1})$ (note that this is always finite by item 4 of Theorem 2.3). Consider

[TABLE]

If in the sequence ${B}_{h}^{N}$ there is no $0\pmod{p}$, then $f(x)$ is a non-zero square modulo $p$ for any $x\in\{1,\dots,N\}$, hence it is a square modulo $p^{s}$ by Lemma 3.1. Assume that there is some $x_{0}\in\{1,\dots,N\}$ such that $h(x_{0})$ is congruent to [math] modulo $p$, so that $b$ is congruent to $-f_{1}x_{0}$ modulo $p$ (there can be at most one such $x_{0}$). In that case, consider instead

[TABLE]

so that $f(x_{0})$ is actually $0\in\mathbb{Z}$, hence a square modulo $p^{s}$, and as before, when $x\neq x_{0}$, $f(x)$ is a non-zero square modulo $p^{s}$. In both cases, ${B}_{f}^{N}$ is a Büchi sequence modulo $p^{s}$.

We now prove that ${B}_{f}$ is a non-trivial Büchi sequence modulo $p^{s}$. It is enough to prove that $f(N+1)$ is not a square modulo $p^{s}$. Indeed, we have $f(N+1)\equiv h(N+1)\pmod{p}$, and the latter is not a square by definition of $h$, so $f(N+1)$ is not even a square modulo $p$. ∎

Next comes the key lemma for having a uniform proof of Theorem 2.2.

Lemma 3.4**.**

Let $p$ be an odd prime and $s\geq 1$. Let $f_{1},f_{2}\in\mathbb{Z}$, with $f_{i}\notin p^{s}\mathbb{Z}$ unless $f_{i}=0$. Assume that not both $f_{1}$ and $f_{2}$ are [math]. Write $t_{1}={\rm ord}_{p}f_{1}$ and $t_{2}={\rm ord}_{p}f_{2}$. For $i\in\{1,2\}$, let $g_{i}$ be such that $f_{i}=p^{t_{i}}g_{i}$ (if $f_{i}=0$, take $g_{i}=0$). Write $m=\min\{t_{1},t_{2}\}$.

If $m$ is even, then we have:

[TABLE]

(where $p^{t_{i}-m}g_{i}$ reads as [math] if $f_{i}=0$). 2. 2.

The sides of the equation in item 1 are infinite if and only if $p=3$, $s=m+1$ and $g_{2}\in 2+3\mathbb{Z}$. 3. 3.

If $m$ is odd, then we have:

[TABLE] 4. 4.

If $t_{2}=\infty$ and $t_{1}$ is odd, then ${\rm opt}(p^{s},0,f_{1})\leq 2$.

Proof.

We first prove items 3 and 4, together with the “$\leq$” inequality in item 1. Let $N\geq 0$ and $f_{0}$ be integers, and write $t_{0}={\rm ord}_{p}f_{0}$ and $f_{0}=p^{t_{0}}g_{0}$ (with $g_{0}=0$ if $f_{0}=0$). Assume that ${B}_{f}^{N}$ is a non-trivial Büchi sequence modulo $p^{s}$, where $f=f_{2}X^{2}+f_{1}X+f_{0}$. In particular, by Proposition 2.1 the polynomial $f$ is not the square of a linear polynomial modulo $p^{s}$, and we have $m\leq t_{0}$, unless $[f_{0}]_{p^{s}}$ is a non-square. If $m>t_{0}$ and $[f_{0}]_{p^{s}}$ is a non-square, we have $N=0$ by Lemma 3.2, so we may assume $m\leq t_{0}$. Write $f=p^{m}g$, where

[TABLE]

We can now complete the proof of item 3. Assume $m$ is odd. In that case, if $[f(n)]_{p^{s}}$ is a square, then $[g(n)]_{p}=[0]_{p}$. If $m=t_{1}<t_{2}$, we have

[TABLE]

hence $g(n)$ can be [math] modulo $p$ for at most one value of $n$, hence $N\leq 1$. If $m=t_{2}\leq t_{1}$, we have

[TABLE]

so $g(n)$ can be [math] modulo $p$ for at most two values of $n$, hence $N\leq 2$.

We now turn to item 1. Assume $m$ is even. Since $f$ is not the square of a linear polynomial modulo $p^{s}$, also $g$ is not the square of a linear polynomial modulo $p^{s-m}$. Moreover, since $t_{0}\geq m=\min\{t_{2},t_{1}\}$, we have

[TABLE]

hence ${B}_{g}^{N}$ is a non-trivial Büchi sequence modulo $p^{s-m}$ by Proposition 2.1, so we have

[TABLE]

We now prove “$\geq$” in item 1 (so in particular, we assume that $m$ is even). First note that the claim is trivial when $t_{2}=0$ (which is the case in particular when $s=1$). Let

[TABLE]

be such that ${B}_{g}^{N}$ is a non-trivial Büchi sequence of length $N={\rm ml}(p^{s-m},p^{t_{2}-m}g_{2},p^{t_{1}-m}g_{1})$ ($N$ may be $\infty$).

Consider

[TABLE]

For any $x\in\{1,\dots,N\}$, since $g(x)$ is a square modulo $p^{s-m}$ and $m$ is even, also $f(x)$ is a square modulo $p^{s}$, so ${B}_{f}$ is a Büchi sequence modulo $p^{s}$.

We now prove that ${B}_{f}^{N}$ is a non-trivial Büchi sequence modulo $p^{s}$. Since ${B}_{g}^{N}$ is a non-trivial Büchi sequence modulo $p^{s-m}$, $g$ is not the square of a linear polynomial modulo $p^{s-m}$, hence also, since $m$ is even, $p^{m}g$ is not the square of a linear polynomial modulo $p^{s}$. Moreover, by Proposition 2.1, either $\min\{t_{1}-m,t_{2}-m\}\leq{\rm ord}_{p}b$, in which case $\min\{t_{1},t_{2}\}\leq{\rm ord}_{p}p^{m}b$, or $[b]_{p^{s-m}}$ is not a square, in which case $[p^{m}b]_{p^{s}}$ is not a square. So ${B}_{f}^{N}$ is a non-trivial Büchi sequence modulo $p^{s}$.

We prove item 2. If $m=t_{1}<t_{2}$, then the right-hand side is finite by Lemma 3.3. Otherwise it is an immediate consequence of Theorem 2.3 applied to modulus $p^{s-m}$ (observe that the only case where ${\rm opt}$ is infinite is in item 5). ∎

Corollary 3.5**.**

Let $p$ be an odd prime and $s\geq 1$. Let $f_{1},f_{2}\in\mathbb{Z}$, with $f_{i}\notin p^{s}\mathbb{Z}$, unless $f_{i}=0$. Write $t_{i}={\rm ord}_{p}f_{i}$ and let $g_{i}$ be such that $f_{i}=p^{t_{i}}g_{i}$ (if $f_{i}=0$, take $g_{i}=0$). We have

[TABLE]

Proof.

First note that the claim is trivial when $t_{2}=0$. If $t_{2}\leq t_{1}$ and $t_{2}$ is even, this is just Lemma 3.4. Assume that $t_{2}>t_{1}$ and $t_{1}$ is even. In particular, since $t_{2}>t_{1}$, $f_{1}$ cannot be [math]. We have

[TABLE]

(recalling the convention that $p^{t_{2}-t_{1}}g_{2}=0$ if $f_{2}=0$), where the first equality comes from Lemma 3.4, and the second equality comes from Lemma 3.3 (which can be applied because $p^{t_{2}-t_{1}}g_{2}$ is not invertible modulo $p$, but $g_{1}$ is invertible modulo $p$ since $f_{1}\neq 0$). ∎

Lemma 3.6**.**

Let $p$ be a prime $\geq 3$ and $s\geq 1$. Let $f_{2}\in\mathbb{Z}$, with $f_{2}\notin p^{s}\mathbb{Z}$ unless $f_{2}=0$. Write $t_{2}={\rm ord}_{p}f_{2}\neq 0$. We have:

If $t_{2}=\infty$, then

[TABLE] 2. 2.

If $t_{2}<\infty$ is odd, then

[TABLE] 3. 3.

If $t_{2}<\infty$ is even, then

[TABLE]

Proof.

For the first item, just note that for any non-zero $b\in\mathbb{Z}$ which is coprime with $p$, the function $f=b^{2}X$ defines a non-trivial Büchi sequence of length $\geq 1$. Indeed, if $f\equiv(g_{1}X+g_{2})^{2}\pmod{p^{s}}$, then $g_{1}^{2}\equiv g_{2}^{2}\equiv 0\pmod{p^{s}}$, hence $[g_{1}]_{p}=[g_{2}]_{p}=0$, but $2g_{1}g_{2}\equiv b^{2}\pmod{p^{s}}$, which contradicts the fact that $b$ is coprime with $p$, and the constant term is [math], hence has order $\geq$ than the order of the other coefficients. We conclude by Proposition 2.1 that it is a non-trivial sequence.

For items 2 and 3, choose $f=f_{2}X^{2}+f_{1}X+f_{0}$ with $f_{1}=1-3f_{2}$ and $f_{0}=2f_{2}-1$, so that $f(1)=0$ and $f(2)=1$ (so they are squares modulo any $p^{s}$). Since $t_{2}\neq 0$, $f_{2}$ is divisible by $p$, hence ${\rm ord}_{p}(f_{1})=0$ is even and $<t_{2}$. Moreover, ${B}_{f}^{2}$ is a non-trivial sequence because, one the one hand we have ${\rm ord}_{p}(f_{0})=0\geq\min\{{\rm ord}_{p}f_{1},{\rm ord}_{p}f_{2}\}$, and on the other hand it is not the square of a linear polynomial modulo $p^{s}$. If it were, then we would have

[TABLE]

which is impossible, since modulo $p$ the right-hand side is a constant polynomial (because $p$ divides $f_{2}$, hence also $g_{1}$), while the left-hand side is a non-constant polynomial since $1-3f_{2}$ is not divisible by $p$. ∎

We conclude this work with the proof of our main theorem.

Proof of Theorem 2.2.

If $f_{2}=0$, then we have (recalling the convention ${\rm opt}(p^{s},0,0)=1$ — see the introduction)

[TABLE]

where the second and third equalities come from item 1 of Lemma 3.6 and Lemma 3.4, and the fourth equality comes from Corollary 3.5.

If $t_{2}<\infty$ is odd, then we have (using again Lemmas 3.6, Lemma 3.4 and Corollary 3.5):

[TABLE]

If $0\neq t_{2}<\infty$ is even, then we have

[TABLE]

∎