Periodic Continued Fractions and Kronecker Symbols

Kurt Girstmair Institut für Mathematik Universität Innsbruck Technikerstr. 13/7 A-6020 Innsbruck Austria

               Kurt.Girstmair@uibk.ac.at

Abstract

We study the Kronecker symbol (\frac{s}{t}) for the sequence of the convergents s/t of a purely periodic continued fraction expansion. Whereas the corresponding sequence of Jacobi symbols is always periodic, it turns out that the sequence of Kronecker symbols may be aperiodic. Our main result describes the period length in the periodic case in terms of the period length of the sequence of Jacobi symbols and gives a necessary and sufficient condition for the occurrence of the aperiodic case.

1 Introduction and main results

Let [a_{0}, a_{1}, a_{2}, …] be the regular continued fraction expansion of an irrational number z. We consider the convergents s_{k}/t_{k}, k ≥ 0, of this expansion. They are defined by the well-known recursion formulas

s_{−1} = 1, t_{−1} = 0, s_{0} = a_{0}, t_{0} = 1,  (1)

and

s_{k} = a_{k}s_{k−1} + s_{k−2}, t_{k} = a_{k}t_{k−1} + t_{k−2},  (2)

for k ≥ 1. Then

s_{k}/t_{k} = [a_{0}, …, a_{k}], k ≥ 0,

is the kth convergent of z; see [4, p. 250]. Note that t_{k} is a positive integer for k ≥ 0.

In two recent papers [2, 3], we investigated the Jacobi symbol (\frac{s_{k}}{t_{k}} ) in the periodic case, i.e., for a quadratic irrational z. Since this symbol is defined only for odd denominators t_{k}, we defined (\frac{s_{k}}{t_{k}} ) = ∗ if t_{k} is even. It turned out that the sequence of Jacobi symbols (\frac{s_{k}}{t_{k}} ), k ≥ 0, is periodic with a period length L = dl, where l is the smallest possible period length of [a_{0}, a_{1}, a_{2}, …] and d is a divisor of 8 or 12. We called this sequence the Jacobi sequence of z.

The natural generalization of the Jacobi symbol (\frac{s}{t}) for arbitrary co-prime integers s, t, t ≥ 1, is the Kronecker symbol. It coincides with the Jacobi symbol if t is odd. If t = 2^{j}t′, where j ≥ 1 and t′ is an odd natural number, one defines

(\frac{s}{t}) = (\frac{s}{2})^{j} (\frac{s}{t′} ) ,

with

(\frac{s}{2}) = \begin{cases}1, & if s ≡ ±1 (mod 8); \\ −1, & if s ≡ ±3 (mod 8); \end{cases}

see [1, p. 28 ff.]. The Kronecker symbol shares many properties with the Jacobi symbol, for instance, the reciprocity law

(\frac{s}{t}) = ε(s′, t′) (\frac{t}{s}) ,  (3)

where s and t are co-prime, s = 2^{j}s′, t = 2^{l}t′ with odd natural numbers s′, t′, and

ε(s′, t′) = \begin{cases}−1, & if s′, t′ both ≡ 3 (mod 4); \\ 1, & otherwise. \end{cases}

So one might think that the periodicity of the Jacobi sequence can be generalized to the corresponding sequence of Kronecker symbols, which we call the Kronecker sequence of z. This, however, is not true, as our main result shows. Furthermore, the Kronecker sequence requires an approach that is considerably different from that of the Jacobi case, as the reader will see in the following sections.

Why did we not settle with less, namely, with the Jacobi sequence? The answer is as follows: simply because our curiosity grew when the difference between the two cases became more and more obvious.

In this paper we restrict ourselves to the purely periodic case since the mixed periodic one seems to be much more difficult. So let z = [\overline{a_{0}, …, a_{l−1}}], where l has been chosen smallest possible. In the paper [2] we have seen that the Jacobi sequence of z is purely periodic with a period length L = dl, d ≥ 1, such that

D_{L} = ( \begin{matrix}s_{L−1} & s_{L−2} \\ t_{L−1} & t_{L−2} \end{matrix} ) ≡ I (mod 4).  (4)

Here I is the 2 × 2 unit matrix and the congruence has to be understood entry-by-entry. Further, we may assume that L is even (in fact, d can be chosen as a divisor of 8 or 12, but we do not require this in what follows, since our periods are not always shortest possible).

In this setting suppose that D_{L} = I + 2^{m}U, m ≥ 2, where not all entries of

U = ( \begin{matrix}x & y \\ u & v \end{matrix} )  (5)

are even. Suppose, further, that u = 2^{e}u′, with e ≥ 0 and u′ odd. A convergent s_{k}/t_{k} of z is called critical with respect to L, if k ≤ L − 1, s_{k} ≡ 3 (mod 4) and t_{k} ≡ 0 (mod 2^{m+e}) (since m ≥ 2, the last-mentioned condition requires t_{k} ≡ 0 (mod 4), of course). Now our main result reads as follows.

Theorem 1. Let the above notation hold, in particular, let z be purely periodic and L be a period length of the Jacobi sequence with the above properties. Suppose, further, that no critical convergent with respect to L exists. Then the Kronecker sequence (\frac{s_{k}}{t_{k}} ), k ≥ 0, is purely periodic with period length L or 2L. If there is, conversely, a critical convergent with respect to L, then the Kronecker sequence is aperiodic.

Remark 2. In Proposition 6 we describe the cases of period length L and 2L of the theorem precisely. As a rule, one finds more examples with periodic Kronecker sequences than with aperiodic ones.

Example 3. For z = [\overline{1, 2, 3}] = (4 + \sqrt{37})/7 we may choose L = 6, m = 2. Here u = 21, so e = 0. There is no critical convergent among s_{0}/t_{0}, …, s_{5}/t_{5}, but the convergent s_{1}/t_{1} = 3/2 has the effect that the Kronecker sequence has only period length 2L = 12.

In the case z = [\overline{1, 2, 5}] = (7 + \sqrt{82})/11 we may choose L = 12, m = 2. Here u is odd, so e = 0. Hence the convergent s_{7}/t_{7} = 975/608 is critical with respect to L and the Kronecker sequence is aperiodic.

An aperiodic example with e > 0 is z = (5 + \sqrt{85})/10 = [\overline{1, 2, 2}], where L = 36 works with m = 3 and e = 1. Therefore, a critical convergent s_{k}/t_{k} must satisfy t_{k} ≡ 0 (mod 16). The convergent s_{6}/t_{6} = 91/64 has this property.

2 The reciprocal Jacobi sequence

An obvious way to generalize the results concerning the Jacobi sequence consists in the generalization of the auxiliary results needed for this purpose. It turns out, however, that this is impossible, as the following example shows. If s/t = [a_{0}, …, a_{k}] is a rational number, then the Jacobi symbol (\frac{s}{t}) (which equals ∗, if t is even) depends only on the residue classes of a_{0}, …, a_{k} mod 4. No result of this kind can hold for the Kronecker symbol. Indeed, let s/t = 3/2^{j} and s′/t′ = 3/(7 · 2^{j}), where j ≥ 3 is odd. We have

s/t = [0, (2^{j} − 2)/3, 1, 2] and s′/t′ = [0, (7 · 2^{j} − 2)/3, 1, 2].

Here (2^{j} −2)/3 ≡ (7·2^{j} −2)/3 (mod 2^{j+1}). The Kronecker symbol, however, takes the values (\frac{s}{t}) = −1 and (\frac{s′}{t′} ) = 1.

Hence our approach to the Kronecker sequence differs from the above strategy of generalizing auxiliary Jacobi results. Instead, we consider the reciprocal Jacobi sequence (\frac{t_{k}}{s_{k}} ),

k ≥ 0 (with (\frac{t_{k}}{s_{k}} ) = ∗ if s_{k} is even). Then we use the reciprocity law (3) in order to obtain the values of the Kronecker symbol in the case where t_{k} is even. For this purpose we need the following proposition.

Proposition 4. Let z = [\overline{a_{0}, …, a_{l−1}}]. Suppose that L = dl, d ≥ 1, is an even period length of the Jacobi sequence of z such that (4) holds. Then the reciprocal Jacobi sequence (\frac{t_{k}}{s_{k}} ), k ≥ 0, is purely periodic with the same period length L.

Proof. From the identity

( \begin{matrix}s_{k+L} & s_{k+L−1} \\ t_{k+L} & t_{k+L−1} \end{matrix} ) = D_{L} · ( \begin{matrix}s_{k} & s_{k−1} \\ t_{k} & t_{k−1} \end{matrix} ) , k ≥ 0,  (6)

(see [2, Eq. (9)] and (4) above) we obtain

s_{j+L} ≡ s_{j} (mod 4), t_{j+L} ≡ t_{j} (mod 4) for all j ≥ −1.  (7)

Hence (\frac{t_{k+L}}{s_{k+L}} ) = (\frac{t_{k}}{s_{k}} ) = ∗ if s_{k} is even, k ≥ 0.

If both s_{k} and t_{k} are odd, we have

(\frac{t_{k+L}}{s_{k+L}} ) = ε(t_{k+L}, s_{k+L}) (\frac{s_{k+L}}{t_{k+L}} ) ,  (8)

by quadratic reciprocity. Now (7) shows ε(t_{k+L}, s_{k+L}) = ε(t_{k}, s_{k}). Since L is a period length of the Jacobi sequence, (\frac{s_{k+L}}{t_{k+L}} ) = (\frac{s_{k}}{t_{k}} ). Accordingly, (8) says

(\frac{t_{k+L}}{s_{k+L}} ) = ε(t_{k}, s_{k}) (\frac{s_{k}}{t_{k}} ) .

Finally, quadratic reciprocity shows (\frac{t_{k+L}}{s_{k+L}} ) = (\frac{t_{k}}{s_{k}} ).

There remains the case t_{k} even, s_{k} odd. Then k ≥ 1, since t_{0} = 1. We use the notation of the proof of [2, Theorem 5] and put s = s_{k−1}, t = t_{k−1}, p = a_{k}, q = 1, m = s_{k}, N = t_{k} and δ = (−1)^{k−1}. Since N is even, t must be odd. Two cases have to be distinguished:

Case 1: s is odd. Then the said theorem yields

(\frac{δt}{s}) (\frac{p}{q}) (\frac{−δN}{m} ) = ε(s, q, m)

with ε(s, q, m) = −1, if two of the numbers s, q, m are ≡ 3 (mod 4), and ε(s, q, m) = 1, otherwise. This gives

(\frac{N}{m}) = (\frac{−δ}{m} ) (\frac{δ}{s}) ε(s, 1, m) (\frac{t}{s}) .

Now ε(s, 1, m) = ε(s, m), and quadratic reciprocity implies

(\frac{N}{m}) = (\frac{−δ}{m} ) (\frac{δ}{s}) ε(s, m)ε(t, s) (\frac{s}{t}) ,

i.e.,

(\frac{t_{k}}{s_{k}} ) = (\frac{−δ}{s_{k}} ) (\frac{δ}{s_{k−1}} ) ε(s_{k−1}, s_{k})ε(t_{k−1}, s_{k−1}) (\frac{s_{k−1}}{t_{k−1}} ) .  (9)

In the same way we obtain

(\frac{t_{k+L}}{s_{k+L}} ) = (\frac{−δ′}{s_{k+L}} ) (\frac{δ′}{s_{k+L−1}} ) ε(s_{k+L−1}, s_{k+L})ε(t_{k+L−1}, s_{k+L−1}) (\frac{s_{k+L−1}}{t_{k+L−1}} ) ,

where δ′ = (−1)^{k+L−1}. However, L is even, so δ′ = δ. Further, all quantities on the right hand side of (9) except the last one depend only on δ and the residue classes of s_{k}, s_{k−1} and t_{k−1} (mod 4), so we may write

(\frac{t_{k+L}}{s_{k+L}} ) = (\frac{−δ}{s_{k}} ) (\frac{δ}{s_{k−1}} ) ε(s_{k−1}, s_{k})ε(t_{k−1}, s_{k−1}) (\frac{s_{k+L−1}}{t_{k+L−1}} ) .

Since L is a period length of the Jacobi sequence, we see that the right hand side of this identity coincides with the right hand side of (9). Thus, (\frac{t_{k+L}}{s_{k+L}} ) = (\frac{t_{k}}{s_{k}} ).

Case 2: s is even. Since t and m are odd, both s + t and m + N are odd. The said theorem gives

(\frac{−δs}{s + t}) (\frac{p}{q}) (\frac{δm}{m + N} ) = ε(s + t, q, m + N)  (10)

Here we use quadratic reciprocity and obtain

(\frac{m}{m + N} ) = ε(m, m + N) (\frac{m + N}{m} ) = ε(m, m + N) (\frac{N}{m}) .  (11)

Similarly,

(\frac{s}{s + t}) = (\frac{−t}{s + t}) = (\frac{−1}{s + t}) (\frac{t}{s + t}) =  (12)
(\frac{−1}{s + t}) ε(t, s + t) (\frac{s + t}{t}) = (\frac{−1}{s + t}) ε(t, s + t) (\frac{s}{t}) .

From (10), (11) and (12) we obtain an expression for (\frac{N}{m}) = (\frac{t_{k}}{s_{k}} ) which depends only on

the residue classes of s_{k}, t_{k}, s_{k−1} t_{k−1} mod 4 and on (\frac{s}{t}) = (\frac{s_{k−1}}{t_{k−1}} ). The same is true for

(\frac{t_{k+L}}{s_{k+L}} ), the residue classes of s_{k+L}, t_{k+L}, s_{k+L−1}, t_{k+L−1} mod 4 and (\frac{s_{k+L−1}}{t_{k+L−1}} ). Since L is a

period length of the Jacobi sequence, we see that (\frac{t_{k+L}}{s_{k+L}} ) equals (\frac{t_{k}}{s_{k}} ).

3 Proof of Theorem 1

As above, let z = [\overline{a_{0}, a_{1}, …, a_{l−1}}] be a purely periodic quadratic irrational, the convergents s_{k}/t_{k} being defined as in (1) and (2). Let L be an even multiple of l such that L is a period length of the Jacobi sequence of z and (4) holds. Again, we write

D_{L} = I + 2^{m}U  (13)

with m ≥ 2, U as in (5) such that not all entries of U are even and u = 2^{e}u′, e ≥ 0, u′ odd. Let k ≥ 0 be such that s_{k} ≡ 3 (mod 4) and t_{k} = 2^{m+f}t′ with −m + 1 ≤ f ≤ e − 1, t′ odd. Note that t_{k} is even but s_{k}/t_{k} is not critical with respect to L in the case k ≤ L − 1.

Lemma 5. In the above setting, let f ≤ e − 2. Then

(\frac{s_{k+L}}{t_{k+L}} ) = (\frac{s_{k}}{t_{k}} )

and t_{k+L} = 2^{m+f}t′′ with t′′ ≡ t′ (mod 4). In the case f = e − 1 we have

(\frac{s_{k+L}}{t_{k+L}} ) = − (\frac{s_{k}}{t_{k}} )

and t_{k+L} = 2^{m+e−1}t′′ with t′′ ≡ t′ + 2 (mod 4).

Proof. From (6), (13) and (5) we obtain

t_{k+L} = 2^{m}us_{k} + t_{k} + 2^{m}vt_{k}.  (14)

Since u = 2^{e}u′ and t_{k} = 2^{m+f}t′, this reads

t_{k+L} = 2^{m+f}(t′ + 2^{e−f}u′s_{k} + 2^{m}vt′).  (15)

If f ≤ e − 2, we obtain t_{k+L} = 2^{m+f}t′′ with t′′ ≡ t′ (mod 4) (observe m ≥ 2). Now the reciprocity law (3) yields

(\frac{s_{k+L}}{t_{k+L}} ) = ε(s_{k+L}, t′′) (\frac{t_{k+L}}{s_{k+L}} ) .

The Kronecker symbol on the right hand side coincides with the Jacobi symbol, since s_{k} and, consequently, s_{k+L} is odd. Moreover, s_{k+L} ≡ s_{k} (mod 4) and t′′ ≡ t′ (mod 4). In addition, the reciprocal Jacobi sequence has the period length L. From this we conclude

(\frac{s_{k+L}}{t_{k+L}} ) = ε(s_{k}, t′) (\frac{t_{k}}{s_{k}} ) .

On applying the reciprocity law (3) again, we have

(\frac{s_{k+L}}{t_{k+L}} ) = (\frac{s_{k}}{t_{k}} ) .  (16)

In the case f = e − 1 we observe that e − f = 1 and u′s_{k} is odd. Accordingly, (15) shows t_{k+L} = 2^{m+e−1}t′′ with t′′ ≡ t′ + 2 (mod 4). Moreover, s_{k+L} ≡ s_{k} ≡ 3 (mod 4), and so ε(s_{k+L}, t′′) = −ε(s_{k}, t′). This produces a sign change on the right hand side of (16).

The periodic case of the Kronecker symbol is contained in the following proposition.

Proposition 6. In the above setting, suppose there are no critical convergents with respect to L. Then the Kronecker sequence is purely periodic with period length L except if there is a convergent s_{k}/t_{k} with k ≤ L − 1, s_{k} ≡ 3 (mod 4) and t_{k} = 2^{m+e−1}t′, t′ odd. In this case the Kronecker sequence is purely periodic with period length 2L.

Proof. We consider an arbitrary convergent s_{k}/t_{k}. If t_{k} is odd, the Kronecker symbol coincides with the Jacobi symbol, which means (\frac{s_{k+L}}{t_{k+L}} ) = (\frac{s_{k}}{t_{k}} ). If t_{k} is even and s_{k} ≡ 1 (mod 4), we have

(\frac{s_{k}}{t_{k}} ) = ε(s_{k}, t′) (\frac{t_{k}}{s_{k}} ) ,

where t′ is the odd part of t_{k}. However, ε(s_{k}, t′) = 1 since s_{k} ≡ 1 (mod 4). Hence we obtain (\frac{s_{k}}{t_{k}} ) = (\frac{t_{k}}{s_{k}} ). In the same way, (\frac{s_{k+L}}{t_{k+L}} ) = (\frac{t_{k+L}}{s_{k+L}} ), because s_{k+L} ≡ s_{k} (mod 4). Now the

periodicity of the reciprocal Jacobi sequence (\frac{t_{k}}{s_{k}} ), k ≥ 0, shows (\frac{s_{k+L}}{t_{k+L}} ) = (\frac{s_{k}}{t_{k}} ).

The case t_{k} even and s_{k} ≡ 3 (mod 4) is contained in Lemma 1. If k ≤ L−1 and f ≤ e−2, we have (\frac{s_{k+dL}}{t_{k+dL}} ) = (\frac{s_{k}}{t_{k}} ) for all natural numbers d. Finally, if k ≤ L − 1 and f = e − 1, we obtain

(\frac{s_{k+dL}}{t_{k+dL}} ) = (−1)^{d} (\frac{s_{k}}{t_{k}} ) .

In this situation the period length is 2L.

Suppose now that k ≤ L − 1 and s_{k}/t_{k} is critical with respect to L. Hence we have s_{k} ≡ 3 (mod 4) and t_{k} ≡ 0 (mod 2^{m+e}). Recall the definition of m and e: by (13), we have D_{L} = I + 2^{m}U, not all entries of U even; and the left lower entry u of U satisfies u = 2^{e}u′, u′ odd. The relation (6) implies D_{2L} = D_{L}^{2}. Therefore, this matrix reads

D_{2L} = I + 2^{m+1}Ũ with Ũ = U + 2^{m−1}U^{2}.  (17)

Since U is as in (5), the left lower entry of U^{2} equals u(x + v), in particular, it is ≡ 0 (mod 2^{e}). Accordingly, the left lower entry ũ of Ũ, i.e., ũ = u + 2^{m−1}u(x + v), is ≡ u (mod 2^{e+1}), since m ≥ 2. Hence ũ = 2^{e}u′′, u′′ odd, with the same exponent e. This means the following. If t_{k} ≡ 0 (mod 2^{m+e+1}), the convergent s_{k}/t_{k} is also critical with respect to 2L. Note that, in this case, m has to be replaced by m + 1 but e remains the same.

Hence there is a number r ≥ 0 such that s_{k}/t_{k} is critical with respect to 2^{r}L but not with respect to 2^{r+1}L. In the following lemma we suppose that r has been chosen in this way. For the sake of simplicity, however, we simply write L instead of 2^{r}L and adopt the other notation connected with D_{L}. Then t_{k} = 2^{m+e}t′, t′ odd.

Lemma 7. In the above setting, let k ≤ L−1. Suppose that s_{k}/t_{k} is critical with respect to L but not critical with respect to 2L. Then s_{k+L}/t_{k+L} is critical with respect to 2L. Moreover, for every d ≥ 1,

(\frac{s_{k+2dL}}{t_{k+2dL}} ) = (−1)^{d} (\frac{s_{k}}{t_{k}} ) .

Proof. As above, we write t_{k} = 2^{m+e}t′, t′ odd. Our situation corresponds to the case f = e in formula (15), and so

t_{k+L} = 2^{m+e}(t′ + u′s_{k} + 2^{m}vt′).

Here, however t′, u′ and s_{k} are odd. Accordingly, t_{k+L} ≡ 0 (mod 2^{m+e+1}), which means that s_{k+L}/t_{k+L} is critical with respect to 2L, as we have seen above.

As in (17) we write D_{2L} = D_{L}^{2} = I + 2^{m+1}Ũ, where the matrix Ũ has the lower entries ũ = 2^{e}u′′, u′′ odd, and ṽ. In this case the analogue of (14) reads

t_{k+2L} = 2^{m+1}ũs_{k} + t_{k} + 2^{m+1}ṽt_{k}.  (18)

If we insert ũ = 2^{e}u′′ and t_{k} = 2^{m+e}t′, we obtain

t_{k+2L} = 2^{m+e}(2u′′s_{k} + t′ + 2^{m+1}ṽt′).

Because u′′ and s_{k} are odd, this yields

t_{k+2L} = 2^{m+e}t′′ with t′′ ≡ t′ + 2 (mod 4).  (19)

As in the proof of Lemma 5 we use the reciprocity law and obtain

(\frac{s_{k+2L}}{t_{k+2L}} ) = ε(s_{k+2L}, t′′) (\frac{t_{k+2L}}{s_{k+2L}} ) .

By (19), ε(s_{k+2L}, t′′) = −ε(s_{k}, t′). Now the periodicity of the reciprocal Jacobi sequence combined with another application of the reciprocity law yields

(\frac{s_{k+2L}}{t_{k+2L}} ) = − (\frac{s_{k}}{t_{k}} ) .

Next formula (18) gives

t_{k+4L} = 2^{m+1}ũs_{k+2L} + t_{k+2L} + 2^{m+1}ṽt_{k+2L}.

On inserting ũ = 2^{e}u′′ and t_{k+2L} = 2^{m+e}t′′, we have

t_{k+4L} = 2^{m+e}(2u′′s_{k+2L} + t′′ + 2^{m+1}ṽt′′).

Since u′′ and s_{k+2L} are odd, this gives

t_{k+4L} = 2^{m+e}t′′′ with t′′′ ≡ t′′ + 2 (mod 4).

Now the above arguments show

(\frac{s_{k+4L}}{t_{k+4L}} ) = − (\frac{s_{k+2L}}{t_{k+2L}} ) = (\frac{s_{k}}{t_{k}} ) .

The general case k + 2dL, d ≥ 1, is settled in the same way by induction.

Let L be such that s_{k}/t_{k} is critical with respect to L. Then there is a number r_{1} ≥ 0 such that s_{k}/t_{k} is critical with respect to 2^{r_{1}}L but not with respect to 2^{r_{1}+1}L. Put k_{1} = k and k_{2} = k + 2^{r_{1}}L. By Lemma 7, s_{k_{2}}/t_{k_{2}} is critical with respect to 2^{r_{1}+1}L. Hence there is a number r_{2} > r_{1} such that s_{k_{2}}/t_{k_{2}} is critical with respect to 2^{r_{2}}L but not with respect to 2^{r_{2}+1}L. In this way we obtain an infinite sequence (k_{j}, r_{j}), j ≥ 1, which is strictly increasing in both arguments such that s_{k_{j}}/t_{k_{j}} is critical with respect to 2^{r_{j}}L but not with respect to 2^{r_{j}+1}L.

Example 8. In the case of the above example (7 + \sqrt{82})/11 = [\overline{1, 2, 5}] the convergent s_{7}/t_{7} is critical with respect to L = 12, but not critical with respect to 2L = 24. Hence we have k_{1} = 7, r_{1} = 0. Since 7 + L = 19, Lemma 7 says that s_{19}/t_{19} is critical with respect to 2L. Since it is not critical with respect to 4L, we have k_{2} = 19, r_{2} = 1. Now 19 + 2L = 43, and s_{43}/t_{43} is critical with respect to 8L, but not with respect to 16L. Hence k_{3} = 43 and r_{3} = 3. Next we have 43 + 8L = 139, s_{139}/t_{139} is critical with respect to 64L, but not with respect to 128L. So we have k_{4} = 139 and r_{4} = 6. Accordingly, the first members of our sequence are (7, 0), (19, 1), (43, 3) and (139, 6).

Proof of Theorem 1. We have only to consider the case that there is a critical convergent with respect to L. Hence we know that there is a sequence (k_{j}, r_{j}) with the above properties.

Suppose that the Kronecker sequence is periodic with period length 2^{i}dL, i ≥ 0, d ≥ 1, d odd. Then there is an integer k_{0} ≥ 0 such that for all k ≥ k_{0}

(\frac{s_{k+2^{i}dL}}{t_{k+2^{i}dL}} ) = (\frac{s_{k}}{t_{k}} ) .

We choose a member (k_{j}, r_{j}) of our sequence in such a way that k_{j} ≥ k_{0} and r_{j} ≥ i − 1. Then 2^{r_{j}+1}dL is a multiple of 2^{i}dL, and so

(\frac{s_{k_{j}+2^{r_{j} +1}dL}}{t_{k_{j}+2^{r_{j} +1}dL}} ) = (\frac{s_{k_{j}}}{t_{k_{j}}} ) ,

by periodicity. By Lemma 7, however,

(\frac{s_{k_{j}+2^{r_{j} +1}dL}}{t_{k_{j}+2^{r_{j} +1}dL}} ) = (−1)^{d} (\frac{s_{k_{j}}}{t_{k_{j}}} ) .

Since d is odd, this is a contradiction.

References

[1] H. Cohen, A Course in Computational Algebraic Number Theory, Springer, 1993.

[2] K. Girstmair, Continued fractions and Jacobi symbols, Int. J. Number Theory 7 (2011), 1543–1555.

[3] K. Girstmair, Periodic continued fractions and Jacobi symbols, Int. J. Number Theory 8 (2012), 1519–1525.

[4] L. K. Hua, Introduction to Number Theory, Springer, 1982.

2010 Mathematics Subject Classification: Primary 11A15; Secondary 11A55. Keywords: Kronecker symbol, convergent, quadratic irrational, periodic continued fraction.

Received May 15 2015; revised versions received August 19 2015; August 20 2015. Published in Journal of Integer Sequences, August 20 2015.

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