1 Introduction

The geometric series

$$\begin{aligned} G_n(x) = 1 + x + x ^ 2 + \cdots + x^{n-1} \end{aligned}$$

(also called geometric progression or GP for short) is an important two-parameter concept used in many branches of mathematics, such as in power series, convergence, telescoping matrix theory [4], number theory [2, 3] and algebraic curves, and has applications in cryptography [1].

For convenience, we shall write \(G_n\) for \(G_n(x)\), when there is no risk of confusion. It is well known that \((x - 1)G_n(x) = x^n - 1\). As such, it is clear that many of the properties of \(G_n(x)\) follow from those of \(x^n - 1\). We shall refer to the latter as the “binomial" of the geometric progression.

When q is a prime power, say \(q=p^e\), the geometric ratio \(G_n(q)\) corresponds to the number of points and of hyperplanes of the projective space \({\mathbb {P}}^{n-1}({\mathbb {F}}_q)\); if it is a prime number, then \(G_n(q)\) is called a projective prime.

The case \(G_2\left( 2^{2^e}\right) \) turns into a Fermat number, whereas \(2^n-1=G_n(2)\) is a Mersènne number. As in these two particular cases, it is conjectured that there exist infinitely many projective primes.

As in the Mersènne numbers, the primality of \(G_n(q)\) implies the primality of n. Indeed, we may use the Product Rule (see (1)) that we will address later to write \(G_n(q)=G_{dt}(q)=G_d(q)G_t(q^d)\), assuming \(n=dt\) is a non-trivial factorization.

Our aim is investigate the fundamental question of when \(G_n(x^p)\) divides \(G_m(x^q)\)—as a polynomial. This four-parameter problem will be referred to as the (n, p, m, q) property.

As always, we shall build on the simpler cases, such as the (n, 1, n, q) and (n, 1, m, q) cases, where \(m = n\) and \(p = 1\), or just when \(p = 1\).

All our results will be over a field \(\mathbb {F}\) with \(char(\mathbb {F}) = 0\). The greatest common divisor and the least common multiple of a and b will be denoted by (a, b) and [a, b], respectively.

We shall need a multitude of preliminary results, which are needed to build our case.

2 Building Blocks

Given integers m and n, let \((m,n) = d\) and suppose that \(n = mq + r\), where \(0\le r < m \le n\). Then,

$$\begin{aligned} x^n - 1 = x^r(x^{mq} - 1) +x^r -1 = (x^m - 1)x^rG_q(x^m) + x^r - 1. \end{aligned}$$

This shows at once that

$$\begin{aligned} m|n \Leftrightarrow x^m - 1|x^n - 1 \Leftrightarrow G_m(x)|G_n(x) \end{aligned}$$

and hence that

$$\begin{aligned} (x^ m- 1, x^n - 1) = x^d - 1 = (x-1)(G_m, G_n). \end{aligned}$$

Consequently,

$$\begin{aligned} G_d = \frac{\textstyle x^d - 1}{\textstyle x - 1} = (G_m, G_n) \end{aligned}$$

and thus

$$\begin{aligned} (G_m,G_n) = 1 \Leftrightarrow (m,n) = 1. \end{aligned}$$

Next, let \(L = [m,n] = lcm(m,n) = \frac{mn}{d}\). We also set \(m = dm'\) and \(n = dn'\) so that \(L = mn' = nm' = m'n'd\).

We now observe that if n|L and m|L, then \( x^n - 1|x^L - 1\) and \( x^m- 1|x^L - 1\). Hence, \([x^m - 1, x^n - 1]|x^L-1|x^{mn}-1\), and thus

$$\begin{aligned} \frac{\textstyle (x^m - 1)( x^n - 1)}{\textstyle (x^d-1)}|x^L-1|x^{mn}-1 \end{aligned}$$

which may be expressed as

$$\begin{aligned} G_m(x)G_n(x)|G_L(x)G_d(x)|G_{mn}(x)G_d(x). \end{aligned}$$

For \(x \ne 1\), we have

$$\begin{aligned} \frac{\textstyle G_{np}(x)}{\textstyle G_p(x)} = \frac{\textstyle x^{np} - 1}{\textstyle x - 1}. \frac{\textstyle x - 1}{\textstyle x^p-1} = \frac{\textstyle x^{np} - 1}{\textstyle x^p- 1} = G_n(x^p), \end{aligned}$$

and thus for all x

$$\begin{aligned} G_{np}(x) = G_p(x)G_n(x^p), \end{aligned}$$
(1)

which we refer to as the Product Rule.

It immediately extends to larger products such as

$$\begin{aligned} G_{abc} = G_aG_{bc}(x^a) = G_aG_b(x^a)G_c(x^{ab}). \end{aligned}$$

A further consequence of the Product Rule is the “q equals one lemma":

Lemma 2.1

(The \(q = 1\) case) The following are equivalent:

  1. (i)

    (n, p, n, 1) holds.

  2. (ii)

    \(G_n(x^p)|G_n(x)\).

  3. (iii)

    \(G_{np} | G_nG_p\).

  4. (iv)

    \(n = 1\) or \(p = 1\).

Proof

The equivalence of (ii)–(iii) follows from the definition and the Product Rule.

If (iii) holds, then using degrees we see that \((np - 1) \le (n-1) + (p-1)\), which tells us that

$$\begin{aligned} (n - 1)(p - 1) \le 0. \end{aligned}$$

Since \(n\ge 1\) and \(p\ge 1\), it follows that (iv) must hold. Lastly, it is clear that (iv) implies (ii). \(\square \)

The following is a key result, which critically depends on the fact that char(\(\mathbb {F}\)) = 0. This will be referred to it as the Linking Lemma with parameter m and links the sub- and superscripts in the two GPs, each of which contains the parameter m.

Lemma 2.2

(Linking Lemma) For any m, n and k,

$$\begin{aligned} (G_{m}(x), G_n(x^{km}) )= 1. \end{aligned}$$

Proof

We begin by noting that \(G_n(1) = n\), which when \(char(\mathbb {F})= 0\) cannot be equal to 0. Now by the remainder theorem

$$\begin{aligned} G_n(x) = (x - 1)Q(x) +G_n(1) \end{aligned}$$

and thus as \(G_n(1) \ne 0\), we conclude that \( (x - 1) \not | ~ G_n(x),\) or

$$\begin{aligned} \left( x - 1, G_n(x) \right) = 1. \end{aligned}$$

Replacing x by \(x^{mk}\) gives \(\left( x^{mk} - 1, G_n(x^{mk})\right) = 1\) and so

$$\begin{aligned} \left( (x - 1)G_m(x) G_k(x^m), G_n(x^{mk}) \right) = 1. \end{aligned}$$

This means that for any m, n and k

$$\begin{aligned} \left( G_m(x), G_n(x^{mk})\right) = 1. \end{aligned}$$

\(\square \)

We use both the Product Rule and the Linking Lemma in the following Basic Lemma, which is a first step in our investigation of \(G_n(x^p) | G_{m}(x^q)\).

Lemma 2.3

((n,1,n,q)) The following are equivalent:

  1. (1)

    \(G_n(x) | G_n(x^q)\) i.e. (n, 1, n, q) holds.

  2. (2)

    \(G_n(x)G_q(x) | G_{qn}(x) \).

  3. (3)

    \((q,n) = 1\).

Proof

From the Product Rule, it is clear that \((1)\Leftrightarrow (2)\).

Let \((q,n) = d\) and \(q = q'd\), \(n = n'd\) and suppose that (1) holds. Then,

$$\begin{aligned} G_n(x) | G_n(x^q) \Rightarrow G_{n'd}(x) | G_n(x^{q'd}) \Rightarrow G_dG_{n'}(x^d) | G_n(x^{q'd}). \end{aligned}$$

By the Linking Lemma, we now get \(G_d = 1\) and thus (3) follows.

Conversely, we always have that

$$\begin{aligned} G_qG_n|G_{qn}G_d \end{aligned}$$

and hence, if \(d = 1\), then (2) follows. \(\square \)

We can immediately extend this to

Lemma 2.4

(Key (n,1,m,q)) The following are equivalent:

  1. (1)

    \(G_n(x) | G_m(x^q)\) i.e. (n,1,m,q) holds.

  2. (2)

    \(G_n(x)G_q(x) | G_{mq}(x)\).

  3. (3)

    \((n,q) = 1\) and n|m.

Proof

The equivalence of (1) and (2) follows again from the Product Rule.

Let \((m,n) = d\) and \(m = m'd\), \(n = n'd\). Also set \((n,q) = e\) and \(n = n''e, q = q''e\). Then, \(G_n(x) = G_e(x)G_{n''}(x^e)|G_m(x^{q''e})\). By the Linking Lemma, with exponent e, we see that \(G_e(x) = 1 \) and thus \(e = (q,n) = 1\). Applying the Basic Lemma, we get \(G_nG_q |G_{nq}\). Combining this with (2), we conclude that

$$\begin{aligned} G_nG_q | (G_{mq}, G_{nq} )= G_{(mq, nq)} = G_{qd}. \end{aligned}$$

This implies that \(G_n | G_{dq}\) and thus n|dq. Since \((n,q) = 1\), it follows that n|d, and we may conclude that \(n = d\) and n|m so that (3) follows.

Conversely, if \((n,q) = 1\), then Lemma 2.3, \(G_nG_q|G_{nq}\) and since n|m, we also have \(G_{nq}|G_{mq}\). Combining these, we arrive at \(G_nG_q|G_{mq}\) giving (2). \(\square \)

3 The Polynomial Ratio

In what follows, we shall need several polynomial results dealing with greatest common divisors. In particular, we recall

Lemma 3.1

Over an Euclidean domain,

  1. 1.

    The gcd Product Rule holds:

    $$\begin{aligned} (ab, cd) = (a,c)(b,d)(a'b', c'd'), \end{aligned}$$

    where \(a' = a/(a,c), c' = c/(a,c), b' = b/(b,d), d' = d/(b,d).\)

  2. 2.
    $$\begin{aligned} (ab, cd) = 1 \text { if and only if } 1 = (a, c) = (a,d) = (b,c) = (b,d). \end{aligned}$$

We now come to a refinement of the four parameters m, n, p and q, indicating the interaction between them.

Given p and q, let \((p,q) = w\) and set \(p = p'w\) and \(q = q'w\), with \((p',q') = 1\).

Consider the rational ratio

$$\begin{aligned} R = \frac{\textstyle G_m(x^q)}{\textstyle G_n(x^p)} = \frac{\textstyle G_m(x^{q'w})}{\textstyle G_n(x^{p'w})} = \frac{\textstyle G_m(y^{q'})}{\textstyle G_n(y^{p'})}, \end{aligned}$$

where \(y = x^w\). Thus, without loss of generality we may assume that \((p,q) = 1\); otherwise, in the final answer replace x by \(x^w\).

We begin by establishing the desired splitting of our four parameters. As such, we define:

$$\begin{aligned} \begin{array}{ccccc} d=(m,n), &{} m = m'd, &{} n = n'd, &{} \text { with }&{} (m',n') = 1\\ f = (m',p),&{} m' = \hat{m}f, &{} p = \hat{p}f, &{}\text { with }&{}(\hat{m},\hat{p}) =1\\ g = (n',q),&{}n' = \bar{n}g,&{}q = \bar{q}g, &{} \text { with } &{} (\bar{n},\bar{q}) = 1\\ h = (\hat{p}, d),&{}\hat{p} = \tilde{p} h, &{} d = \tilde{d} h,&{}\text { with } &{} ( \tilde{p},\tilde{d}) = 1\\ t = (\bar{q}, d),&{}\bar{q} = q'' t,&{}d = d'' t,&{} \text { with }&{} (q'', d'') = 1. \end{array} \end{aligned}$$

Further, we set \(r = \hat{m}\bar{q}\) and \(s = \hat{p}\bar{n}\).

Because \((m',n') = 1 = (p,q)\), we know that \(e = (m'q, n'p) = (m',p)(n',q) = fg\).

We also observe that

$$\begin{aligned} (s,r) = (\hat{p}\cdot \bar{n}, \hat{m}\bar{q}) = 1, \end{aligned}$$

because all four partial gcds equal one, i.e. \((\hat{p}, \bar{q}) = 1 = (\hat{m}, \bar{n}) = (\hat{p},\hat{m}) = (\bar{n}, \bar{q})\).

From the Product Rule, we know that

$$\begin{aligned} R \text { is a polynomial} \Leftrightarrow G_n(x^p)|G_m(x^q) \Leftrightarrow \frac{\textstyle G_{np}}{\textstyle G_p} | \frac{\textstyle G_{mq}}{\textstyle G_q}\Leftrightarrow G_qG_{np}|G_pG_{mq}. \end{aligned}$$

Now \(np = (de)( \hat{p}\bar{n}) = (de)s\) and \(mq = (de) \hat{m}\bar{q} = (de)r\) and hence

$$\begin{aligned} R \text { is a polynomial} \Leftrightarrow G_qG_{de}G_{\hat{p}\bar{n}}(x^{de}) |G_pG_{de}G_{\hat{m}\bar{q}}(x^{de}) \Leftrightarrow G_qG_s(x^{de}) |G_pG_r(x^{de}). \end{aligned}$$

Because \((p,q) = 1 = (r,s)\), we know that \((G_p(x), G_q(x)) = 1 = (G_r(x^{de}), G_s(x^{de}))\). And thus R will be a polynomial if and only if both of the following conditions hold:

$$\begin{aligned} \text {(I) }~~G_q(x) |G_r(x^{de}) \, \text { and } \, \text {(II) }~~ G_s(x^{de}) |G_p. \end{aligned}$$

Let us now examine these two conditions.

Turning to condition (I), we have \(q = g \bar{q}\) and \(r = \hat{m}\bar{q}\), and thus

$$\begin{aligned} G_q(x) |G_r(x^{de}) \Leftrightarrow G_g(x) G_{\bar{q}}(x^g)| G_{\bar{q}}(x^{de})G_{\hat{m}}(x^{de\bar{q}}). \end{aligned}$$

Since g divides de, we can use the Linking Lemma to conclude that \(G_g\) is coprime to both factors of the RHS. As such, we must have \(G_g = 1\), and thus \(g = 1\). This means that \( n' = \bar{n}\) and \(q = \bar{q}\).

We are left with

$$\begin{aligned} G_{\bar{q}} | G_{\bar{q}} (x^{de}) G_{\hat{m}}(x^{de\bar{q}}). \end{aligned}$$

Again, the Linking Lemma implies that

$$\begin{aligned} ( G_{\bar{q}}, G_{\hat{m}}(x^{de\bar{q}})) = 1 \end{aligned}$$

which leaves us with

$$\begin{aligned} G_{\bar{q}} | G_{\bar{q}} (x^{de}). \end{aligned}$$

Using the Basic (n, 1, n, q) Lemma, we arrive at \((\bar{q},de) = (q, de) = 1\). This shows that

$$\begin{aligned} t = (q,d) = 1 \end{aligned}$$
(2)

in addition to \((q,e) = (q,f) = 1\).

Turning to the second condition (II) with \(e = f\), we see that splitting \(s = \hat{p}\bar{n}\) and \(p = \hat{p}f\), we deduce that

$$\begin{aligned} G_s(x^{df}) |G_p \Leftrightarrow G_{\hat{p}}(x^{df})\cdot G_{\bar{n}}(x^{de\hat{p}})| G_f\cdot G_{\hat{p}}(x^f). \end{aligned}$$

Because \(f|df|df\hat{p}\) and \(\hat{p}|df\hat{p}\), we may conclude that

$$\begin{aligned} G_{\bar{n}}(x^{de\hat{p}}) = 1 \end{aligned}$$

and thus we must have

$$\begin{aligned} \bar{n} = 1. \end{aligned}$$

This ensures that \(n' = \bar{n}\cdot g = 1\) and hence \(n = d = (m,n)\) or

$$\begin{aligned} n|m. \end{aligned}$$

We are left with

$$\begin{aligned} G_{\hat{p}}(x^{de})| G_{\hat{p}}(x^f). \end{aligned}$$

Comparing degrees

$$\begin{aligned} (\hat{p} - 1)df\le (\hat{p} - 1)f \end{aligned}$$

or by using the “q equals one Lemma”, we see that either \(\hat{p} = 1\) or \(d = 1\). In the latter case, we get \(n = n'd = 1\cdot 1 = 1\), which is excluded.

On the other hand, when \(\hat{p}= 1\), \(p = f = (m', p)\) so that \(p|m' = \frac{m}{d} = \frac{m}{n} \).

Combining these results with (2), we see that if R is a polynomial, then n|m, \(p| \frac{\textstyle m}{\textstyle n}\) and \((n,q) = 1\).

Conversely, suppose n|m, \(p| \frac{\textstyle m}{\textstyle n}\) and \((q,n) = 1\).

The latter shows that \((q, pn) = (q,p)(q,n) = 1\). Next, let \(m = m'n \), \(m' = p\) and \(m = npw\). As np divides npw, and \((np,q) = 1\), we see by the Key (n, 1, m, q) Lemma that \(G_{np} |G_{npw}(x^q)\). Hence,

$$\begin{aligned} G_{np} | G_p G_{pnw}(x^q) ~\text { or }~ G_n(x^p) | G_m(x^q). \end{aligned}$$

We have proven

Theorem 3.1

$$\begin{aligned} \frac{G_m\left( x^q\right) }{G_n\left( x^p\right) }\text { is a polynomial } \end{aligned}$$

if and only if

$$\begin{aligned} n|m,\, \, p|\frac{m}{n},\, \, (n, q)=1. \end{aligned}$$

4 Remarks

The above establishes when the ratio R will be a polynomial. However, it does not tell us what the actual polynomial is or when it will again be a GP. Also, the ratio question is a first step towards the computation of the gcd of two GPs. These topics will involve geometric series of the form \(G_n(-x)\) with negative arguments and will be addressed in a later examination.

We close with a couple of non-trivial examples.

  1. 1.

    The (6, 3, 18, 5) case, with \(n = 6, p = 3, m = 18, q = 5\). In this case, it is clear that 3|(18/6) and \((5,6) =1\). The GPs are \(G_{18}(x^5) = x^{85} + x^{80} + x^{75} + x^{70} + x^{65} + x^{60} + x^{55} + x^{50} + x^{45} + x^{40} + x^{35} + x^{30} + x^{25} + x^{20} + x^{15} + x^{10} + x^{5} + 1 \), and \( G_6(x^3) = x^{15} + x^{12} + x^{9} + x^{6} + x^{3} + 1.\) The quotient \(R = \frac{\textstyle G_{18}(x^5)}{\textstyle G_6(x^3)}\) equals \(x^{70} - x^{67} + x^{65} - x^{62} + x^{60} - x^{57} + x^{55} + x^{50} - x^{49} + x^{45} - x^{44} + x^{40} - x^{39} + x^{35} - x^{31} + x^{30} - x^{26} + x^{25} - x^{21} + x^{20} + x^{15} - x^{13} + x^{10} - x^{8} + x^{5} - x^{3} + 1\).

  2. 2.

    The (4, 1, 4, 3) case, with \(n=4, p=1, m=4, q=3.\) The GPs are \(G_4(x^3) = x^9 + x^6 + x^3 + 1\) and \(G_4(x) = x^3 + x^2 + x + 1\). This time, \(R = \frac{\textstyle G_4(x^3)}{\textstyle G_4(x)} = x^{6} - x^{5} + x^{3} - x + 1 = G_3(-x^2)G_3(-x)\).