A Congruence Relation of the Catalan-Mersenne Numbers1

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Indian J. Pure Appl. Math., 49(3): 521-526, September 2018 °c Indian National Science Academy DOI: 10.1007/s13226-018-0281-8

A CONGRUENCE RELATION OF THE CATALAN-MERSENNE NUMBERS1

Ick Sun Eum

Department of Mathematics Education, Dongguk University-Gyeongju 123 Dongdae-ro, Gyeongju, Gyeongbuk, 38066, South Korea e-mail: [email protected]

(Received 16 August 2017; accepted 6 October 2017)

The Catalan-Mersenne numbers cn are double Mersenne numbers deﬁned by c0 = 2 and

cn−1 cn = 2 − 1 for positive integers n. We prove a certain congruence relation of the Catalan- Mersenne numbers.

Key words : Mersenne number; congruences.

1. INTRODUCTION

Let c0 = 2. For a positive integer n, the n-th Catalan-Mersenne number is deﬁned by

cn−1 cn = 2 − 1.

127 In 1876 Lucas proved that c4 = 2 − 1 is also a prime by using his primality test, known as the Lucas-Lehmer test [2, 3]. In 1876 and 1891 Catalan [1] noticed that the Mersenne numbers

2 c1 = 2 − 1 = 3, 3 c2 = 2 − 1 = 7, 7 c3 = 2 − 1 = 127, 127 c4 = 2 − 1 = 170141183460469231731687303715884105727 were all primes and he conjectured that the numbers c1, c2, c3, c4, c5, ... are all primes “up to a certain limit”. But it is not known whether or not the 5th Catalan-Mersenne number c5 is prime because it is too huge.

1This work was supported by the Dongguk University Research Fund of 2017 and the National Research Foundation of Korea(NRF) grant funded by the Korea government(MSIT) (No. NRF-2017R1C1B5017567). 522 ICK SUN EUM

m Fermat’s little theorem implies that if m is prime, then the Mersenne number Mm = 2 − 1 satisﬁes

Mm ≡ 1 (mod m). (1) The converse does not hold in general. In this paper, we shall prove that the congruence (1) also holds for all double Mersenne numbers. Moreover we shall show that the Catalan-Mersenne numbers satisfy stronger congruence relation than (1).

2. CONGRUENCE RELATIONS

Mn The numbers of the form MMn = 2 − 1 are called double Mersenne numbers. From the deﬁni- tion, the Catalan-Mersenne numbers are a special case of the double Mersenne numbers. First, we introduce the following proposition.

Proposition 2.1 — Let p be a positive integer.

(i) Mp is a prime only if the exponent p is a prime.

(ii) Suppose that p is an odd prime. Then every prime divisor of Mp is of the form 2pk + 1.

Moreover, Mp is also of the same form, that is, Mp = 2ph + 1 for some positive integer h.

PROOF : See [4, Chapter 7, §7.3]. 2

Remark 2.2 : (i) When the exponent p is not a prime, Proposition 2.1 (ii) does not hold in general. For instance, we have 26 − 1 = 63 = 2 · 31 + 1 which is not of the form 2 · 6h + 1.

(ii) However, we will see in the following proposition that a part of Proposition 2.1 (ii) still valid for all double Mersenne numbers.

Mp,n Proposition 2.3 — Let p be an odd prime. Let Mp,1 = Mp and Mp,n+1 = 2 − 1 for positive integers n. Then we have

Mp,n+1 ≡ 1 (mod Mp,n). (2)

PROOF : To show this, we will use induction on n. Since p is an odd prime, by Proposition 2.1

(ii), Mp,1 = Mp = 2ph + 1 for some positive integer h. We note that

Mp,1 2ph+1 p 2h Mp,2 − 1 = 2 − 2 = 2 − 2 = 2((2 ) − 1) ³ ´ = 2(2p − 1) (2p)2h−1 + (2p)2h−2 + ··· + (2p)1 + 1 ³ ´ p 2h−1 p 2h−2 p 1 = 2Mp,1 (2 ) + (2 ) + ··· + (2 ) + 1

≡ 0 (mod Mp,1). A CONGRUENCE RELATION OF THE CATALAN-MERSENNE NUMBERS 523

Now, we suppose that the congruence (2) holds for n. Then there is a positive integer k such that

Mp,n+1 = 2Mp,nk + 1. From the induction hypothesis, we have ³ ´ Mp,n+1 2Mp,nk+1 Mp,n 2k Mp,n+2 − 1 = 2 − 2 = 2 − 2 = 2 (2 ) − 1 ³ ´ = 2(2Mp,n − 1) (2Mp,n )2k−1 + (2Mp,n )2k−2 + ··· + (2Mp,n )1 + 1 ³ ´ Mp,n 2k−1 Mp,n 2k−2 Mp,n 1 = 2Mp,n+1 (2 ) + (2 ) + ··· + (2 ) + 1

≡ 0 (mod Mp,n+1).

Hence, the congruence (2) also holds for n + 1 and this proves our proposition. 2

Example 2.4 : Let p = 11. Then the Mersenne number 211 − 1 = 23 · 89 is composite. But, by Proposition 2.3, we get 11 22 −1 − 1 ≡ 1 (mod 211 − 1).

Moreover, the Catalan-Mersenne numbers satisfy the following stronger congruence than (2).

Theorem 2.5 — Let m0 = 1. For each positive integer n there exists a positive integer mn such that Yn n+1−i n n−1 2 mn−1|mn and cn+1 = 2mn ci + 1 = 2c1 · c2 ··· cn−1 · cn · mn + 1. (3) i=1

In other words, n n−1 2 cn+1 ≡ 1 (mod 2c1 · c2 ··· cn−1 · cn). (4)

PROOF : We will prove (3) by using induction on n. First, one can see that

c1 3 c2 = 2 − 1 = 2 − 1 = 7 = 2 · 3 · 1 + 1 = 2 · c1 · 1 + 1,

c2 7 2 2 c3 = 2 − 1 = 2 − 1 = 127 = 2 · 3 · 7 · 1 + 1 = 2 · c1 · c2 · 1 + 1.

This shows that m1 = 1, m2 = 1, m0|m1 and m1|m2. Thus (3) holds for n = 1, 2.

Now we suppose that (3) holds for n and n + 1. Then, by induction hypothesis, there are positive integers mn and mn+1 such that

Yn nY+1 n+1−i n+2−i mn|mn+1, cn+1 = 2mn ci + 1 and cn+2 = 2mn+1 ci + 1. (5) i=1 i=1 524 ICK SUN EUM

Since mn|mn+1, we have (cn+1 − 1) | (cn+2 − 1) and cn+2 − 1 mn+1 = c1 · c2 ··· cn+1 · ∈ Z. cn+1 − 1 mn From the deﬁnition, we derive that

cn+2−1 cn+3 − 1 = 2(2 − 1) cn+2−1 (cn+1−1) = 2(2 cn+1−1 − 1) c −1 n+2 −1 cn+1−1 X ¡ ¢k = 2(2cn+1−1 − 1) · 2cn+1−1 k=0 c −1 n+2 −1 cn+1X−1 ¡ ¢ cn+1−1 k = (cn+2 − 1) · 2 . k=0

nY+1 n+2−i Therefore, we have (cn+2 − 1)|(cn+3 − 1). Since cn+2 − 1 = 2mn+1 ci , it remains to i=1 show that c −1 n+2 −1 cn+1X−1 ¡ ¢ cn+1−1 k 2 ≡ 0 (mod c1 · c2 ··· cn+1 · cn+2). k=0 By (5), we have c + 1 nY+1 nY+1 2cn+1−1 = n+2 = m cn+2−i + 1 ≡ 1 (mod c ), 2 n+1 i i i=1 i=1 which implies that cn+2−1 cn+2−1 c −1 −1 c −1 −1 n+1X ¡ ¢ n+1X c − 1 m nY+1 cn+1−1 k n+2 n+1 2 ≡ 1 ≡ ≡ c1 · c2 ··· cn+1 · ≡ 0 (mod ci). cn+1 − 1 mn k=0 k=0 i=1 On the other hand,

cn+2−1 cn+3 − 1 = 2(2 − 1) cn+2−1 cn+1 c = 2((2 ) n+1 − 1) since cn+1|cn+2 − 1 by (5) c −1 n+2 −1 cnX+1 = 2(2cn+1 − 1) · (2cn+1 )k k=0 c −1 n+2 −1 cnX+1 cn+1 k = 2cn+2 · (2 ) k=0 ≡ 0 (mod cn+2). A CONGRUENCE RELATION OF THE CATALAN-MERSENNE NUMBERS 525

Since gcd(cn+2, cn+2 − 1) = 1, cn+2 divides

c −1 n+2 −1 cn+1−1 c − 1 X ¡ ¢k n+3 = 2cn+1−1 . cn+2 − 1 k=0

Also, since gcd(cn+2, ci) = 1 for 1 ≤ i ≤ n + 1 by (5), we can conclude that

c −1 n+2 −1 cn+1X−1 ¡ ¢ cn+1−1 k c1 · c2 ··· cn+1 · cn+2 divides 2 . k=0

Therefore, there exists a positive integer me n+2 so that

c −1 n+2 −1 cn+1X−1 ¡ ¢ cn+1−1 k 2 = c1 · c2 ··· cn+1 · cn+2 · me n+2. k=0

Put mn+2 = mn+1me n+2. Combining the above results, we obtain that

c −1 n+2 −1 cn+1X−1 ¡ ¢ cn+1−1 k cn+3 − 1 = (cn+2 − 1) · 2 Ã k=0 ! nY+1 n+2−i = 2mn+1 ci · (c1 · c2 ··· cn+1 · cn+2 · me n+2) i=1 nY+2 n+3−i = 2mn+1me n+2 ci i=1 nY+2 n+3−i = 2mn+2 ci . i=1

By construction, mn+1|mn+2 and (3) also holds for n + 2. This proves our assertion. 2

Example 2.6 : For the case when n = 1, 2 and 3, we have

c1 3 c2 = 2 − 1 = 2 − 1 = 7 = 2 · 3 · 1 + 1 = 2 · c1 · 1 + 1 ≡ 1 (mod 2c1),

c2 7 2 2 2 c3 = 2 − 1 = 2 − 1 = 127 = 2 · 3 · 7 · 1 + 1 = 2 · c1 · c2 · 1 + 1 ≡ 1 (mod 2c1c2),

c3 127 c4 = 2 − 1 = 2 − 1 = 170141183460469231731687303715884105726 + 1 = 2 · 33 · 72 · 127 · 19 · 43 · 73 · 337 · 5419 · 92737 · 649657 · 77158673929 + 1 3 2 3 2 = 2 · c1 · c2 · c3 · m3 + 1 ≡ 1 (mod 2c1c2c3). 526 ICK SUN EUM

2127−1 51217599719369681875006054625051616349 We note that c5 = 2 −1 > 10 . Thus c5 −1 is too huge to ﬁnd its prime factorization. However, Theorem 2.5 tells us that

4 3 2 c5 ≡ 1 (mod 2c1c2c3c4) ≡ 1 (mod 2 · 34 · 73 · 1272 · 170141183460469231731687303715884105727).

Remark 2.7 : Let us consider the Mersenne number 25 −1 = 31 which is a prime. It is well-known that 5 22 −1 − 1 = 231 − 1 = 2147483647 is also a Mersenne prime. However, one can check that

5 22 −1 − 1 = 231 − 1 = 2147483647 ≡ 1 (mod 31) ≡ 22 6≡ 1 (mod 52).

Hence the congruence (4) does not hold in general.

REFERENCES

1. L. E. Dickson, History of the theory of numbers, Vol. 1: Divisibility and primality, New York, Dover (2005).

2. D. H. Lehmer, An extended theory of Lucas’ functions, Ann. of Math. (2), 31(3) (1930), 419-448.

3. E.´ Lucas, Theorie´ des fonctions numeriques´ simplement periodiques,´ Amer. J. Math., 1 (1878), 184-240, 289-321.

4. K. H. Rosen, Elementary number theory and its applications, sixth edition, Pearson Addison Wesley (2011).