On Worley s theorem in Diophantine approximations

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1 Annales Mathematicae et Informaticae 35 (008) pp On Worley s theorem in Diophantine approximations Andrej Dujella a, Bernadin Irahimpašić a Department of Mathematics, University of Zagre Pedagogical Faculty, University of Bihać Sumitted 11 March 008; Accepted 17 Octoer 008 Astract In this paper we prove several results on connection etween continued fractions and rational approximations of the form α a/ < /, for a positive integer. Keywords: Continued fractions MSC: Primary 11A55, 11J70; Secondary 11D Introduction The classical Legendre s theorem in Diophantine approximations states that if a real numer α and a rational numer a (we will always assume that 1), satisfy the inequality α a < 1, (1.1) then a is a convergent of the continued fraction expansion of α = [a 0; a 1,...]. This result has een extended y Fatou [3] (see also [5, p.16]), who showed that if α a < 1, then a = pm q m or pm+1±pm q m+1±q m, where pm q m denotes the m-th convergent of α. In 1981, Worley [1] generalized these results to the inequality α a <, where is an aritrary positive real numer. Worley s result was slightly improved in [1]. The first author was supported y the Ministry of Science, Education and Sports, Repulic of Croatia, grant

2 6 A. Dujella, B. Irahimpašić Theorem 1.1 (Worley [1], Dujella [1]). Let α e a real numer and let a and e coprime nonzero integers, satisfying the inequality α a <, (1.) where is a positive real numer. Then (a, ) = (rp m+1 ± sp m, rq m+1 ± sq m ), for some m 1 and nonnegative integers r and s such that rs <. The original result of Worley [1, Theorem 1] contains three types of solutions to the inequality (1.). Two types correspond to two possile choices for signs + and in (rp m+1 ± sp m, rq m+1 ± sq m ), while [1, Theorem 1] shows that the third type (corresponding to the case a m+ = 1) can e omitted. In Section 3 we will show that Theorem 1.1 is sharp, in the sense that the condition rs < cannot e replaced y rs < ( ε) for any ε > 0. However, it appears that the coefficients r and s show different ehavior. So, improvements of Theorem 1.1 are possile if we allow nonsymmetric conditions on r and s. Indeed, already the paper of Worley [1] contains an important contriution in that direction. Theorem 1. (Worley [1], Theorem ). If α is an irrational numer, 1 and a is a rational approximation to α (in reduced form) for which the inequality (1.) holds, then either a pm is a convergent q m to α or a has one of the following forms: (i) a = r > s and rs <, or rpm+1+spm rq m+1+sq m r s and rs < + r a m+, (ii) a = s < t and st <, or spm+1 tpm sq m+1 tq m s t and st ( 1 s) t <, where r, s and t are positive integers. Since the fraction a/ is in reduced form, it is clear that in the statements of Theorems 1.1 and 1. we may assume that gcd(r, s) = 1 and gcd(s, t) = 1. Worley [1, Corollary, p.06] also gave the explicit version of his result for = : α a < implies a = pm q m, pm+1±pm q m+1±q m, pm+1±pm q m+1±q m, 3pm+1+pm 3q m+1+q m, pm+1±pm q m+1±q m or pm+1 3pm q m+1 3q m. This result for = has een applied for solving some Diophantine equations. In [7], it was applied to the prolem of finding positive integers a and such that (a + )/(a + 1) is an integer, and in [] it was used for solving the family of Thue inequalities x 4 4cx 3 y + (6c + )x y + 4cxy + y 4 6c + 4. On the other hand, Theorem 1.1 has applications in cryptography, too. Namely, in [1], a modification of Verheul and van Tilorg variant of Wiener s attac ([10, 11]) on RSA cryptosystem with small secret exponent has een descried, which is ased on Theorem 1.1.

3 On Worley s theorem in Diophantine approximations 63 We will extend Worley s wor and give explicit and sharp versions of Theorems 1.1 and 1. for = 3, 4, 5,..., 1. We will list the pairs (r, s) which appear in the expression of solutions of (1.) in the form (a, ) = (rp m+1 ± sp m, rq m+1 ± sq m ), and we will show y explicit examples that all pairs from the list are indeed necessary. We hope that our results will also find applications on Diophantine prolems, and in Section 4 we will present such an application. In such applications, it is especially of interest to have smallest possile list of pairs (r, s). It is certainly possile to extend our result for > 1. However, already our results mae it possile to reveal certain patterns, and they also suffice for our Diophantine applications.. Explicit versions of Worley s theorem We start y few details from the proof of Theorem 1.1 in [1], which will e useful in our future arguments. In particular, we will explain how the integer m appearing in the statement of Theorem 1.1 can e found. We assume that α < a, since the other case is completely analogous. Let m e the largest odd integer satisfying α < a p m q m. If a > p1 q 1, we tae m = 1, following the convention that p 1 = 1, q 1 = 0. Since p m+1 q m p m q m+1 = 1, the numers r and s defined y are integers, and since pm+1 q m+1 < a maximality of m, we find that a = rp m+1 + sp m, sa m+ r q m+ = From (.1) we immediately have and we can derive the inequality = rq m+1 + sq m pm q m, we have that r 0 and s > 0. From the p m+ a α q m+ < a <. (.1) a m+ > r s, (.) r sra m+ + a m+ > 0 (.3) (see [1, proof of Theorem 1] for details, and note also that (.3) is exactly the inequality from Theorem 1. (i) - the second case). Let us define a positive integer t y t = sa m+ r. Then we have a = rp m+1 + sp m = sp m+ tp m+1, = rq m+1 + sq m = sq m+ tq m+1,

4 64 A. Dujella, B. Irahimpašić and s and t satisfy analogs of (.) and (.3): a m+ > t s, (.4) t sta m+ + a m+ > 0. (.5) If r > t, i.e. rs > st, then we will represent a and in terms of s and t (which corresponds to sign in Theorem 1.1). Let us consider now the case = 3. Hence, we are considering the inequality α a < 3. (.6) By Theorem 1.1, we have that (a, ) = (rp m+1 + sp m, rq m+1 + sq m ) or (sp m+ tp m+1, sq m+ tq m+1 ), where rs < 6, st < 6, gcd(r, s) = 1 and gcd(s, t) = 1. However, the inequalities (.3) and (.5) for r = 1, resp. t = 1, show that the pairs (r, s) = (1, 4), (1, 5) and (s, t) = (4, 1), (5, 1) can e omitted. Therefore, we proved Proposition.1. If a real numer α and a rational numer a satisfy the inequality (.6), then a = rp m+1 + sp m, where rq m+1 + sq m (r, s) R 3 = {(0, 1), (1, 1), (1, ), (1, 3), (, 1), (3, 1), (4, 1), (5, 1)}, or a = sp m+ tp m+1 sq m+ tq m+1, where (for an integer m 1). (s, t) T 3 = {(1, 1), (, 1), (3, 1), (1, ), (1, 3), (1, 4), (1, 5)} Our next aim is to show that Proposition.1 is sharp, i.e. that if we omit any of the pairs (r, s) or (s, t) appearing in Proposition.1, the statement of the proposition will no longer e valid. More precisely, if we omit a pair (r, s ) R 3, satisfying (.6), ut such then there exist a real numer α and a rational numer a that a cannot e represented in the form a = rpm+1+spm rq m+1+sq m nor a = spm+ tpm+1 sq m+ tq m+1, where m 1, (r, s) R 3 \ {(r, s )}, (s, t) T 3 (and similarly for an omitted pair (s, t ) T 3 ). We will show that y giving explicit examples for each pair. Although we have found many such examples of different form, in the next tale we give numers α of the form d, where d is a non-square positive integer. α a m r s t α a m r s t

5 On Worley s theorem in Diophantine approximations 65 For example, consider α = 8 = [, 1, 4]. Its rational approximation 3 8 (the forth row of the tale) satisfies < 3 8. The convergents of 8 are 1, 3 1, 14 5, 17 6, 8 9, 99 35, ,.... The only representation of the fraction 3 8 in the form rpm+1+spm rq m+1+sq m, (r, s) R 3 or spm+ tpm+1 sq m+ tq m+1, (s, t) T 3 is 3 8 = = 1 p +3 p 1 1 q +3 q 1, which shows that the pair (1, 3) cannot e omitted from the set R 3. Proposition.. Let {4, 5, 6, 7, 8, 9, 10, 11, 1}. If a real numer α and a rational numer a satisfy the inequality (1.), then a = rp m+1 + sp m rq m+1 + sq m, where (r, s) R = R 1 R, or a = sp m+ tp m+1, where (s, t) T = T 1 T (for an sq m+ tq m+1 integer m 1), where the sets R and T are given in the following tale. Moreover, if any of the elements in sets R or T is omitted, the statement will no longer e valid. R T 4 {(1, 4),(3, ),(6, 1),(7,1)} {(4,1), (,3), (1,6),(1, 7)} 5 {(1, 5),(, 3),(8, 1),(9,1)} {(5,1), (3,), (1,8),(1, 9)} 6 {(1, 6),(5,), (10, 1),(11, 1)} {(6, 1),(, 5),(1, 10),(1, 11)} 7 {(1, 7), (,5),(4, 3),(1, 1),(13, 1)} {(7, 1),(5,), (3,4), (1,1), (1,13)} 8 {(1, 8), (3,4),(7, ),(14, 1),(15, 1)} {(8, 1),(4,3), (,7), (1,14), (1,15)} 9 {(1, 9),(5,3), (16, 1),(17, 1)} {(9, 1),(3, 5),(1, 16),(1, 17)} 10 {(1,10), (9, ),(18, 1),(19, 1)} {(10, 1), (,9),(1, 18), (1,19)} 11 {(1, 11), (,7), (3,5),(0, 1), (1,1)} {(11, 1),(7, ),(5,3), (1,0), (1,1)} 1 {(1, 1),(5, 4),(7, 3), {(1, 1), (4,5), (3,7), (11, ),(, 1), (3,1)} (,11), (1,), (1,3)} Proof. By Theorem 1.1, we have to consider only pairs of nonnegative integers (r, s) and (s, t) satisfying rs <, st <, gcd(r, s) = 1 and gcd(s, t) = 1. Furthermore, as in the case = 3, it follows directly from the inequalities (.3) and (.5) for r = 1, resp. t = 1, that the pairs (r, s) = (1, s) and (s, t) = (s, 1) with s + 1 can e omitted. Similarly, for r = or 3, resp. t = or 3, we can exclude the pairs (r, s) = (, s) and (s, t) = (s, ) with s +, and the pairs (r, s) = (3, s) and (s, t) = (s, 3) with s Now we show that all remaining possile pairs which are not listed in the statement of Proposition. can e replaced with other pairs with smaller products rs, resp. st. We give details only for pairs (r, s), since the proof for pairs (s, t) is completely analogous (using the inequalities (.4) and (.5), instead of (.) and (.3)). Consider the case = 4 and (r, s) = (, 3). By (.3), we otain a m+ <. Thus, the pair (r, s) = (, 3) can appear only for a m+ = 1. However, in that case we have t = sa m+ r = 1, and therefore the (r, s) = (, 3) can e replaced y the pair (s, t) = (3, 1). Analogously we can show that for = 7 the pair (r, s) = (3, 4) can e replaced y (s, t) = (4, 1), for = 8, 9, 10 the pair (r, s) = (3, 5) can e replaced y (s, t) = (5, ), while for = 11, 1 the pair (r, s) = (4, 5) can e replaced y (s, t) = (5, 1).

6 66 A. Dujella, B. Irahimpašić We have only three remaining pairs to consider: the pair (r, s) = (5, 3) for = 8 and the pairs (r, s) = (5, 4) and (r, s) = (7, 3) for = 11. For (r, s) = (5, 3) and = 8, from (.) and (.3) we otain 5 3 < a m+ < 5 7, and therefore we have two possiilities: a m+ = or a m+ = 3. If a m+ =, we can replace (r, s) = (5, 3) y (s, t) = (3, 1), while if a m+ = 3, we can replace it y (s, t) = (3, 4). Similar approach wors for two pairs with = 11. For (r, s) = (5, 4), from (.) and (.3) we otain 5 4 < a m+ < 5 9, which implies a m+ =. Then we have t = 3 and the pair (r, s) = (5, 4) can e replaced y the pair (s, t) = (4, 3). For (r, s) = (7, 3) we otain 7 3 < a m+ < 49 10, which yields a m+ = 3 or a m+ = 4. If a m+ = 3, we can replace (r, s) = (7, 3) y (s, t) = (3, ), while if a m+ = 4, we can replace it y (s, t) = (3, 5). It remains to show that all pairs listed in the statement of the proposition are indeed necessary (they cannot e omitted). This is shown y the examples from the following tales: = 4 α a m r s t = 5 α a m r s t = 6 α a m r s t = 7 α a m r s t = 8 α a m r s t = 9 α a m r s t

7 On Worley s theorem in Diophantine approximations 67 = 10 α a m r s t = 11 α a m r s t = 1 α a m r s t For example, tae the first row for = 1, i.e. α = 384 = [61, 1, 60, 1, 1] and its rational approximation , which satisfies < The convergents of 384 are 61 1, 6 1, , , , , ,.... The only representation of the fraction in the form rpm+1+spm rq m+1+sq m, (r, s) R 1 or sp m+ tp m+1 sq m+ tq m+1, (s, t) T 1 is = = 1 p4+1 p3 1 q 4+1 q 3, which shows that the pair (1, 1) cannot e omitted from the set R Cases r = 1, s = 1 and t = 1 The results from the previous section suggest that there are some patterns in pairs (r, s) and (s, t) which appear in representations (a, ) = (rp m+1 + sp m, rq m+1 +sq m ) and (a, ) = (sp m+ tp m+1, sq m+ tq m+1 ) of solutions of inequality (1.). In particular, these patterns are easy to recognize for pairs of the form (r, s) = (r, 1) or (1, s), and (s, t) = (s, 1) or (1, t). In this section we will prove that the results on these pairs, already proved for 1, are valid in general. These facts will allow us to show that the inequality rs < in Theorem 1.1 is sharp. We will assume that is a positive integer. From Theorem 1.1 it directly follows that among the pairs of the form (r, 1), only pairs where r 1 can appear. Similarly, for pairs (1, t) we have t 1. On the other hand, from (.3) and (.5) it follows that for pairs (1, s) we have s, and for pairs (s, 1) we have s. These results follow also from Theorem 1.. We will show that all these pairs that do not contradict Theorem 1. can indeed appear.

8 68 A. Dujella, B. Irahimpašić Let α m = [a m ; a m+1, a m+,...] and 1 β m the convention that β 1 = 0. Then for a = rpm+1+spm rq m+1+sq m, we have = qm 1 q m = [a m 1, a m,..., a 1 ], with α a (rqm+1 = + sq m ) αm+pm+1+pm α m+q m+1+q m (rp m+1 + sp m ) = sαm+ r (rqm+1+sqm) α m+q m+1+q m = sαm+ r (r+sβm+) α m++β m+. (3.1) We start with the pairs of the form (r, 1). Let us consider the numer α = Its continued fraction expansion has the form = [; 4] (see e.g. [8, p.97]). Tae first m = 1, i.e. consider the rational numer a defined y a = r p p 1 = r + 1 = + 1 r q q 1 r r. Hence, a = r+1 and = r. We claim that for r 1, α a < ( holds. By (3.1), this is equivalent to 1 r α 1 )r <. For m 1 we have α m = [4, 4,...] < Thus, it suffices to chec that 4r (16 + 1)r > 0, which is clearly satisfied for r 1. More precisely, this is satisfied for r less than > 1. We can proceed similarly for m 0. The only difference is that 4 < 1 β m+ = [4,...,4] < Hence, y (3.1), we otain that it suffices to chec that for r 1, ( ) r r < holds. But this condition is equivalent to ( )r ( )r + ( ) > 0, which holds for r less than 3 4, so it certainly holds for r 1. The same example α = can e used to handle the pairs (s, t) = (1, t). The relation (3.1) can e reformulated in terms of s and t = sa m+ r: ) α a = (t + s α m+3 s t+ s α m+3 α m++β m+. (3.) Now, for m = 1 we are considering the rational numer a = s p 1 t p 0 = t s q 1 t q 0 4 t = t. By (3.), the condition α a < leads to 16 t 64 3 t > 0. Similarly, for m 0, we otain the condition 8 t (3 3 +)t > 0. It is easy to see that oth conditions are satisfied for t 1.

9 On Worley s theorem in Diophantine approximations 69 For pairs of the form (1, s) and (s, 1) we use α of the form α = x 1, where the integer x will e specified latter (if necessary). For x, we have the following continued fraction expansion x 1 = [x 1; 1, x ] (see e.g. [8, p.97]). Let us consider the pairs of the form (r, s) = (1, s). We tae m = 1 and define the rational numer a = 1 p 0 + s p 1 = x 1 + s. 1 q 0 + s q 1 1 Hence, a = x 1 + s and = 1, and for s, α a < (x 1 + s) (x 1) = s holds. The same result for pairs (r, s) = (1, ) holds also if m 1 is odd and if x is sufficiently large. Indeed, from (3.1) we otain the condition ( ( ) ) ( ) 1 + x 1 x 1 + <, x 1 which is satisfied for x +5. Finally, consider the pairs of the form (s, t) = (s, 1) for s. Tae m = 1 and define the rational numer a = s p 1 1 p 0 = sx x + 1 = x + 1 s q 1 1 q 0 s 1 s 1. Hence, a = sx x + 1 and = s 1. We have x 1 > p q = x 1 x 1. Thus, α a < 1 s x 1, and we otain the condition 1 s x 1 < If we choose x to e greater than + 1, then we have s the inequality (s 1) 1 s 1 (s 1). (3.3) ( 1), while for ( 1) 1 1 = 1 ( 1) holds, and we showed x 1 < 1 that for such x s the condition (3.3) is fulfilled. Again, the analogous result for pairs (s, t) = (, 1) holds for all odd m 1, ut x has to e larger than in the case m = 1. Namely, the relation (3.) yields the condition ( 1 + ) ( ) 1 x 1 + <, x

10 70 A. Dujella, B. Irahimpašić which is satisfied for x +6. Our results for the pairs (r, s) = ( 1, 1) and (s, t) = (1, 1) (with α = 4 + 1) immediately imply the following result which shows that Theorem 1.1 is sharp. Proposition 3.1. For each ε > 0 there exist a positive integer, a real numer α and a rational numer a, such that α a <, and a cannot e represented in the form a = rpm+1±spm rq m+1±sq m, for m 1 and nonnegative integers r and s such that rs < ( ε). Proof. Tae > 1 ε, α = and e.g. a = ( 1)+1 1. Then α a <. If m = 1, then r = 1, s = 1, t = +1, and thus rs = 1 > ε = ( ε), while st = + 1. If m 0, then from s = p m+1 + aq m+1 it follows that s a p1 q q1 1 = + 1, and therefore rs + 1 and st A Diophantine application In [], Dujella and Jadrijević considered the Thue inequality x 4 4cx 3 y + (6c + )x y + 4cxy 3 + y 4 6c + 4, where c 3 is an integer. In this section we will assume that c 5, since the cases c = 3 and c = 4 require somewhat different details. Using the method of Tzanais [9], they showed that solving the Thue equation x 4 4cx 3 y+(6c + )x y +4cxy 3 + y 4 = µ, µ Z \ {0}, reduces to solving the system of Pellian equations (c + 1)U cv = µ (4.1) (c )U cz = µ, (4.) where U = x + y, V = x + xy y and Z = x + 4xy + y. If suffices to find solutions of the system (4.1) and (4.) which satisfy the condition gcd(u, V, Z) = 1. Then gcd(u, V ) = 1, and gcd(u, Z) = 1 or, since 4V +Z = 5U. It is clear that the solutions of the system (4.1) and (4.) induce good rational approximations of the corresponding quadratic irrationals. More precisely, from [, Lemma 4] we have the inequalities given in the following lemma. Lemma 4.1. Let c 5 e an integer. All positive integer solutions (U, V, Z) of the system of Pellian equations (4.1) and (4.) satisfy c + 1 V < c U U (4.3)

11 On Worley s theorem in Diophantine approximations 71 c Z c U < 6c + 4 U c (c ) < 9 U. (4.4) Using the result of Worley [1, Corollary, p. 06], in [, Proposition ] the authors proved that if µ is an integer such that µ 6c + 4 and that the equation (4.1) has a solution in relatively prime integers U and V, then µ {1, c, c + 1, 6c + 1, 6c + 4}. Analysing the system (4.1) and (4.), and using the properties of convergents of c+1 c, they were ale to show that the system has no solutions for µ = c, c + 1, 6c + 1. Applying results from the previous sections to the equation (4.), we will present here a new proof of that result, ased on the precise information on µ s for which (4.) has a solution in integers U and Z such that gcd(u, Z) {1, }. The simple continued fraction expansion of a quadratic irrational α = e+ d f is periodic. This expansion can e otained using the following algorithm. Multiplying the numerator and the denominator y f, if necessary, we may assume that f (d e ). Let s 0 = e, t 0 = f and a n = sn+ d, s n+1 = a n t n s n, t n+1 = d s n+1 t n for n 0 (4.5) t n (see [6, Chapter 7.7]). If (s j, t j ) = (s, t ) for j <, then Applying this algorithm to α = [a 0 ;..., a j 1, a j,..., a 1 ]. c c = c(c ) c, we find that c = [0; 1, c, ]. c According to our results (Proposition. for = 9), applied to α = c c, all solutions of (4.) have the form Z/U = (rp m+1 + sp m )/(rq m+1 + sq m ) an index m 1 and integers r and s. For the determination of the corresponding µ s, we use the following result (see [, Lemma 1]): Lemma 4.. Let αβ e a positive integer which is not a perfect square, and let p m /q m denotes the mth convergent of continued fraction expansion of α β. Let the sequences (s m ) and (t m ) e defined y (4.5) for the quadratic irrational αβ β. Then α(rq m+1 + sq m ) β(rp m+1 + sp m ) = ( 1) m (s t m+1 + rss m+ r t m+ ). (4.6) c Since the period of the continued fraction expansion of c is equal to, according to Lemma 4., we have to consider only the fractions rpm+1+spm rq m+1+sq m for m = 1 and m =. By checing all possiilities, we otain the following result.

12 7 A. Dujella, B. Irahimpašić Proposition 4.3. Let µ e an integer such that µ 6c + 4 and that the equation (c )U cz = µ has a solution in integers U and Z such that gcd(u, Z) = 1 or. (i) If c 15 is odd, then µ M 1 = {1, 4, c, 4c + 1, 6c + 4, c + 4, 4c + 9, 6c + 16}. Furthermore, if c = 5, 11, 13, then µ M 1 { 8c + 5}; if c = 9, then µ M 1 { 8c + 5, 10c + 36}; if c = 7, then µ M 1 { 8c + 5, 10c + 36, 1c + 49}. (ii) Let M = M 1 M, where { M = 11 c + 36, 9 c + 5, 7 c + 16, 5 c + 9, 3 c + 4, 1 c + 1, 1 c, 3 c + 1, 5 c + 4, 7 } c + 9. If c 108 is even, then µ M { 9 11 c + 16, c + 5}. For even c with 6 c 106, we have µ M M (c), where M (c) can e given explicitly, as in the case (i). E.g. { M (6) = 1 } c + 5, 10c + 36, 8c + 5, 15 c Comparing the set {1, c, c+1, 6c+1, 6c+4} from [, Proposition ] with the sets appearing in Proposition 4.3, we otain the desired conclusion. Corollary 4.4. Let c 5 e an integer. If the system (4.1) and (4.) has a solution with µ 6c + 4 in integers U, V and Z such that gcd(u, V, Z) = 1, then µ = 1 or µ = 6c + 4. Acnowledgements. The authors would lie to than the referee for valuale remars and suggestions. References [1] Dujella, A., Continued fractions and RSA with small secret exponents, Tatra Mt. Math. Pul., 9 (004) [] Dujella, A., Jadrijević, B., A family of quartic Thue inequalities, Acta Arith., 111 (004) [3] Fatou, P., Sur l approximation des incommenurales et les series trigonometriques, C. R. Acad. Sci. (Paris), 139 (1904)

13 On Worley s theorem in Diophantine approximations 73 [4] Kosma, J.F., On continued fraction, Simon Stevin, 9 (1951/5) [5] Lang, S., Introduction to Diophantine Approximations, Addison-Wesley, Reading, [6] Niven, I., Zucerman, H.S., Montgomery, H.L., An Introduction to the Theory of Numers, John Wiley, New Yor, [7] Osgood, C.F., Luca, F., Walsh, P.G., Diophantine approximations and a prolem from the 1988 IMO, Rocy Mountain J. Math., 36 (006) [8] Sierpińsi, W., Elementary Theory of Numers, PWN, Warszawa; North-Holland, Amsterdam, [9] Tzanais, N., Explicit solution of a class of quartic Thue equations, Acta Arith. 64 (1993) [10] Verheul, E.R., van Tilorg, H.C.A., Cryptanalysis of less short RSA secret exponents, Appl. Algera Engrg. Comm. Computing, 8 (1997) [11] Wiener, M.J., Cryptanalysis of short RSA secret exponents, IEEE Trans. Inform. Theory, 36 (1990) [1] Worley, R.T., Estimating α p/q, J. Austral. Math. Soc. Ser. A, 31 (1981) Andrej Dujella Department of Mathematics, University of Zagre Bijeniča cesta 30, Zagre, Croatia duje@math.hr Bernadin Irahimpašić Pedagogical Faculty, University of Bihać Džanića mahala 36, Bihać, Bosnia and Herzegovina ernadin@ih.net.a