IJISET - International Journal of Innovative Science, Engineering & Technology, Vol. 6 Issue 12, December 2019 ISSN (Online) 2348 – 7968 | Impact Factor (2019) – 6.248 www.ijiset.com

FOUNDATIONS ON SEMI-PRIMITIVE ROOTS

ALIREZA ALIZADEH MOGHADDAM ∗

Abstract. Z∗ Let n denoted as a multiplicative group of inte- ∈ Z∗ gers modulo n. Any element a n is said to be a semi- primitive root if the set of H = {1, a, a2, ..., a(φ(n)/2)−1} is Z∗ the cyclic subgroup of n. On this article, we have dis- Z∗ cussed about some cyclic subgroup of the set of n which is generated by a semi-primitive root modulo n and we have reviewed some properties, Finally we have also studied the solution of some modular arithmetic equations by using of semi-primitive roots.

1. Introduction Definition 1.1. Let n be a positive integer, all of the integer numbers between 1 and n which are coprime to n form a group with multiplication modulo n as the Z∗ binary operation, this group denoted by n and is called the multiplicative group of integers modulo n. the cardinality of this group denoted by φ(n), where φ(n) is the Euler’s phi- function. ∈ Z∗ φ(n) ≡ Note 1.1. It is obvious that for each a n , Euler’s theorem states that a 1(mod n), and it’s order which is denoted by ordn(a), is the smallest positive integer number k such that ak ≡ 1(mod n). Therefore a is said to be a primitive root modulo n ( i.e. primitive root) if the order of a modulo n is equal to φ(n). Z∗ Note 1.2. It is well known that n as a cyclic multiplicative group usually has a Z∗ k k primitive root, also the n is cyclic if and only if n is equals to 1, 2, 4, p or 2p , where p is an odd prime number and k ≥ 1.

Definition 1.2. Let n be any positive integer number, a ∈ Zn is denoted as a Z∗ semi-primitive root modulo n, if the set of H is a cyclic subgroup of n as the following: φ(n) − ∗ { 2 2 1} ≤ Z H = 1, a, a , ..., a n ∈ Z∗ 2 ∈ Z∗ It is obvious that if a n be a primitive root , then a n is a semi-primitive root.

2010 Mathematics Subject Classification. 11A07, 11A05. Key words and phrases. Multiplicative group of integers modulo n, Primitive roots, Semi- primitive roots. ∗Department of Mathematics, Faculty of Science, University of Yasouj, Yasouj, IRAN. 1

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2 A. R. ALIZADEH M.

Z∗ Also all of non-cyclic groups as the form of n which are contain some semi-primitive roots are classified by the following Theorem. Z∗ Theorem 1.1. Let n be the multiplicative group of integers modulo n that does Z∗ not possess any primitive roots, then n has a semi-primitive root if and only if n k k k1 k2 k1 k2 is equals to 2 (k > 2), 4p1 1, p1 p2 or 2p1 p2 , where p1 and p2 are odd prime k1 k2 ≥ integers satisfying (φ(p1 ), φ(p2 )) and k1, k2 1. Proof. See reference [4] □ Note 1.3. It is shown that in [3] that 3 is a semi-primitive root modulo 2k for any Z∗ integer k greater than 2 and 2k , k > 2 can be represented as Z∗ { i k−2} 2k = 3 mod n : i = 1, ..., 2 . Z∗ ∼ × Theorem 1.2. Suppose n = C2 Cφ(n)/2, then there exists a semi-primitive root ∈ Z∗ h n such that φ(n) Z∗ = {hi mod n : i = 1, ..., }. n 2 Proof. See reference [4] □ ∈ Z∗ Z∗ Definition 1.3. The a n is said to be a good semi-primitive root ”GSP ” if n can be expressed as Z∗ × − n =< h > < 1 > Z∗ Theorem 1.3. If n be a non-cyclic group contains some semi-primitive roots, Z∗ φ(n) then n has exactly 2φ( 2 ) incongruent ”GSP ” roots. Z∗ × − Proof. According by Theorem ”1.2”, n =< h > < 1 > for a semi-primitive ∈ Z∗ ∈ Z∗ i − i root h n. In other words, any element a n can be expressed a = h or h , φ(n) where i = 1, 2, ..., 2 . For the case of a = hi,

i ordn(h) φ(n)/2 ordn(h ) = = (ordn(h), i) (φ(n)/2, i)

Where ordn(a) indicates the order of a modulo n. This implies that φ(n) φ(n) ord (hi) = ⇐⇒ ( , i) = 1. n 2 2 Since h is a ”GP S” root modulo n, it is also clear that (hi)j ≠ −1 for all integers φ(n) Z∗ j. Therefore they said that there are φ( 2 ) incongruent ”GSP ” roots in n in the form of hi. φ(n) Now they shown that there are also φ( 2 ) incongruent ”GSP ” roots modulo n Z∗ − i − i in n in the form of h . More precisely, h is a ”GSP ” roots modulo n if and φ(n) only if ( 2 , i) = 1. They first note that i − i i 2ordn(h ) ( φ(n) ) ordn( h ) = [2, ordn(h )] = i = ( ) . (2, ordn(h )) φ(n) φ(n)/2 2 , i 2, ((φ(n)/2,i),i) ( ) − i φ(n) φ(n) φ(n)/2 Then ordn( h ) = 2 if and only if ( 2 , i) 2, (φ(n)/2,i) = 2. In other words, φ(n) (1) ( φ(n) , i) = 1 and ( φ(n) , 2) = 2 or ord (−hi) = ⇐⇒ 2 2 n 2 φ(n) φ(n) (2) ( 2 , i) = 2 and ( 2 , 2) = 1

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FOUNDATIONS ON SEMI-PRIMITIVE ROOTS 3

φ(n) Since 2 is even for the cases under their consideration, the first case is simply φ(n) φ(n) − i j ̸≡ − ( 2 , i) = 1. If ( 2 , i) = 1 , ( h ) 1 for all integers j. φ(n) − i j ≡ − Suppose that ( 2 , i) = 1 and ( h ) 1 for an integer j, then j must be odd integer , since h is a ”GP S” root modulo n, which it gives that hij ≡ −1 , φ(n) | equivalently 2 ij. φ(n) φ(n) φ(n) | Since ( 2 , i) = 1 and 2 is even, therefore 2 j and hence j must be an even integer, which is leading to contradiction. φ(n) φ(n) φ(n) For the second case, if ( 2 , i) = 2 and ( 4 , 2) = 1, then 4 is an odd integer i φ(n) φ(n) 1 and i is an even integer. Then (−h ) 4 = (−1)(h 2 ) 2 ≡ −1 mod n, which means that any semi-primitive root modulo n in the second case is not a ”GSP ”. − i φ(n) □ This implies that the h is a ”GSP ” root modulo n if and only if ( 2 , i) = 1 Z∗ Note 1.4. By the style of proving in the last theorem, it is shown that if n is φ(n) a noncyclic group possessing semi-primitive roots and 4 is an odd integer, then Z∗ φ(n) φ(n) n has φ( 4 ) more semi-primitive roots in addition to 2φ( 2 )”GSP ” roots. Now we want to show that the way of finding a ”GSP ” root in a non-cyclic multiplicative group.

Theorem 1.4. Let m1 and m2 be coprime integers possessing primitive roots a and b respectively. If (φ(m1), φ(m2)) = 2 , there exist some semi-primitive roots modulo n = m1m2 and the solution to the system of linear congruences: x ≡ a(mod m ) (1) 1 x ≡ −b(mod m2) is a ”GSP” root modulo n. Proof. By Chinese Remainder Theorem it is ensured that the system in (1) has a unique answer modulo n, say x0. Now by the hypotheses since a and b are primitive roots and (m1, m2) = 1 , it concludes that [φ(m1),φ(m2)] ≡ x0 1(mod n) k ≡ k ≡ In addition the equation of x0 1(mod n) equivalent to a 1(mod m1) and k (−b) ≡ 1(mod m2). Since a is a primitive root modulo m1 and φ(m1)|k and (φ(m1), φ(m2)) = 2 so k k the k must be an even integer. Therefore (−b) = b ≡ 1(mod m2), equivalently φ(m2)|k. Hence: φ(m )φ(m ) φ(n) ord (x ) = [φ(m ), φ(m )] = 1 2 = n 0 1 2 2 2

This implies that x0 is a semi-primitive root modulo n. φ(n)/4 ̸≡ − Now if x0 1(mod n) , then x0 will be a ”GSP ” root modulo n. φ(n)/4 ≡ − ≡ For this, let x0 1(mod m1m2) , then by the first congruence x a(mod m1) we have φ(m1) φ(m2) φ(m2) −1 ≡ (a 2 ) 2 ≡ (−1) 2 (mod m1)

φ(m2) and therefore 2 is an odd integer. In the meantime, the second congruence x ≡ −b(mod m2) gives

φ(m2) φ(m1) φ(m2) φ(m1) φ(m1) φ(m1) −1 ≡ ((−1) 2 ) 2 (b 2 ) 2 ≡ (−1) 2 (−1) 2 (mod m2)

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4 A. R. ALIZADEH M.

Which is a contradiction □ Z∗ By the last two Theorems and their proofs we conclude that, if n is a non-cyclic Z∗ φ(n) group possessing semi-primitive roots then n has exactly 2φ( 2 ) (i.e. φ(φ(n))) incongruent ”GSP ” roots and also the number of semi-primitive roots is either φ(n) φ(n) φ(φ(n)) or φ(φ(n)) + φ( 4 ) according as 4 is even or odd.

2. Some Properties of Semi-primitive Roots In this section, it has been produced some key results for the semi-primitive roots Z∗ and ”GSP ” roots in n At the first step for producing of the number of semi-primitive roots for different positive values of n, we define { ∈ Z∗ | } S(n) = g n g is a semi-primitive root modulo n So { φ(φ(n)) + φ( φ(n) if φ(n) is odd | | 4 4 S(n) = φ(n) φ(φ(n)) if 4 is even Where cardinality of S(n) is denoted by |S(n)|. By the following proposition it is shown that the number of semi-primitive roots in Z∗ n is always greater than 2. Z∗ Theorem 2.1. Let n be a non-cyclic group possessing a semi-primitive root. Then the number of semi-primitive roots is at least 3 φ(n) Proof. The number of semi-primitive roots is given by φ(φ(n)) + φ( 4 or φ(n) φ(φ(n)) according as 4 is odd or even. As φ(n) > 2 , so φ(φ(n)) is even. Therefore number of semi-primitive roots is greater than 1. Consider the smallest possible case φ(φ(n)) = 2. Then φ(n) = 3, 4 or 6. The φ(n) only possibility is φ(n) = 4 , which implies that 4 is odd. So the number semi-primitive roots is greater than 2. □ Z∗ Theorem 2.2. Let n be a non-cyclic group possessing a semi-primitive root. Then Z∗ 3 n has exactly 3 semi-primitive roots if and only if n = 2 or 12 Proof. See reference [4] □ Z∗ Theorem 2.3. Let n be a non-cyclic group possessing a semi-primitive root. Then the number of semi-primitive roots is always an even number except for n = 23, 12. Proof. See reference [4] □ Z∗ Theorem 2.4. Let n be a non-cyclic group possessing a semi-primitive root. Let 2p, where p is an odd prime number be the number of semi-primitive roots, then p = 3 and n = 21, 28, 42. Proof. See reference [4] □ Z∗ Corollary 2.1. It is easy to see that if n is a non-cyclic group possessing semi- primitive root and if number of semi-primitive roots is of the form 2kp, where p is φ(n) ≤ ≤ an odd prime then for 4 is even,0 k 31 and p = 3. ∈ Z∗ Theorem 2.5. Every a n is a semi-primitive root modulo n if and only if φ(n) ordn(a) = 2 .

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FOUNDATIONS ON SEMI-PRIMITIVE ROOTS 5

Proof. (⇒): Let a be a semi-primitive root modulo n, then the elements of 0 1 φ(n) −1 0 1 = a , a , ..., a 2 produced a cyclic group modulo n, in addition since a = 1 φ(n) φ(n) ≤ ≤ − k ̸≡ 2 ≡ and for each 1 k 2 1 and we have a 1(mod n) and a 1(mod n) so any of their two elements doesn’t congruent together, therefore: φ(n) ord (a) = n 2 ⇐ φ(n) ( ): Let ordn(a) = 2 since we have the following set: 0 1 φ(n) −1 a , a , ..., a 2 such that for every k , n ̸ |ak, now it is enough to showing the all of their elements are distinct modulo n. For this, if this assume is not true , so there exists two elements ≤ ≤ φ(n) − i ≡ j j i j−i i, j such that: 0 i < j 2 1 and a a (mod n), then: a = a a and n ̸ |ai so ai can be deleted of two sides. therefore we have: 1 ≡ aj−i(mod n) ≤ − φ(n) − φ(n) □ and since 1 j i < 2 1 then this is contradict by ordn(a) = 2 . Corollary 2.2. Let a, n ∈ Z+ be such that n ̸ |a and let a2 = b then: | φ(n) i) ordn(b) 2 t ii) If b ≡ 1(mod n), then ordn(b)|t. + u ordn(b) iii) For every u ∈ Z , ordn(b ) = . (ordn(b),u) Proof. i) By Euler Theorem we usually have, aφ(n) ≡ 1(mod n), then: φ(n) φ(n) 2 ≡ | (b) 1(n), therefore ordn(b) 2 . ii) Let ordn(b) = k, by division algorithm we have: t = kq + r , 0 ≤ r ≤ k Then 1 = (b)t ≡ (b)kq+r = (bk)qbr ≡ 1qbr(mod n) Since k is the smallest integer number, such that satisfying (b)k ≡ 1(mod n), so the relation of (b)r ≡ 1(mod n) , 0 ≤ r ≤ k it is concluded that r = 0 and therefore k|t

iii) By letting k = ordn(b) and m = (u, ordn(b)) = (u, k) and by (ii) we have: (bu)t = but ≡ 1(mod n) ⇐⇒ k|ut But this relation equivalent to: k |ut m m Since k u ( , ) = 1 m m then, k |t m

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6 A. R. ALIZADEH M.

So we have shown that k (bu)t ≡ 1(mod n) ⇐⇒ ( )|t m k Since the ( m ) is the smallest integer t such that satisfying: (bu)t ≡ 1(mod n) therefore we have: k ord (b) = n m □ Z∗ Corollary 2.3. It is easy to see that if n is a non-cyclic group possessing semi- primitive root and if number of semi-primitive roots is of the form 2kp, where p is φ(n) ≤ ≤ an odd prime then for 4 is even,0 k 31 and p = 3. Lemma 2.1. If n ̸ |a , a2 = b and (b)∗ is inverted arithmetic of (b), then: ∗ ordn(b) = ordn(b) k ≡ Proof. Let k = ordn (b), since (b) 1(mod n), we have: ((b)∗)k = 1 × (bk)∗ ≡ (bk)(bk)∗ ≡ 1(mod n) therefore: ∗ ordn(b )|k ∗ Now by letting ordn(b ) = t since we have: (b)t ≡ ((b)∗)t(b)t ≡ ((b)∗b)t ≡ 1t ≡ 1(mod n) ∗ Then the ordn(b) = k divid the t = ordn(b ) (i.e. k|t), Therefore: t = k □

+ Theorem 2.6. Let n ∈ Z and the integers a1 , a2 be such that n ̸ |a1 , n ̸ |a2 , 2 2 a1 = b1 , a2 = b2 and ordn(b1 ) = k1 , ordn(b2 ) = k2 which are (k1 , k2 ) = 1, then:

ordn(b1 b2 ) = k1 k2 Proof. Since k k k k k k k k ((b b )) 1 2 = (b 1 ) 2 (b 2 ) 1 ≡ 1 1 1 2 ≡ 1(mod n) 1 2 1 2 Then | ordn((b1 b2 )) k1 k2 Now since t ≡ ⇒ t t ≡ ((b1 b2 )) 1(mod n) = (b1 ) (b2 ) 1(mod n) t t Therefore (b2 ) is an arithmetic inverse of (b1 ) . Therefore by Lemma ”2.2” we have: t t ordn(b1 ) = ordn(b2 ) then: k k 1 = 2 (t, k1 ) (t, k2 )

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FOUNDATIONS ON SEMI-PRIMITIVE ROOTS 7

Since (k1 , k2 ) = 1 in addition the k1 and k2 haven’t any common factor greater than 1, then we have: k k 1 = 2 = 1 (t, k1 ) (t, k2 ) | | Then by (t, k2 ) = k1 we have: k1 t, as a similar way we have: k2 t, and again since

(k1 , k2 ) = 1, then we have: | k1 k2 t By all of the last concludes we have: | k1 k2 ordn((b1 b2 )) And in the other wise it is obvious that | ordn((b1 b2 )) k1 k2 Hence:

ordn((b1 b2 )) = k1 k2 □

Theorem 2.7. Let a be a semi-primitive root of modulo n, then: φ(n) ai ≡ ai(mod n) ⇐⇒ i ≡ j(mod ) 2 0 1 φ(n) −1 Proof. Let a be a semi-primitive root of modulo n, then the set of {a , a , ..., a 2 } is a cyclic multiplicative group generated by a which is the subset of the set of all reduced residue set modulo n such that incongruent together modulo n Now if i < j and (a)i ≡ (a)j(mod n) then: (a)j−i ≡ 1(mod n) φ(n) therefore by ordn (a) = 2 it is implied that: φ(n) |j − i 2 φ(n) | − φ(n) Conversely: If 2 j i then j = i + k( 2 ) and so: j (i+k( φ(n) )) i ( φ(n) )k i i (a) = (a) 2 = (a) ((a) 2 ) ≡ (a) × 1 ≡ (a) (mod n) Therefore the proof is completed. □ Theorem 2.8. Let n, b ∈ Z+ such that: n ̸ |b, if k be a positive integer and φ(n) k ≡ s = (k, 2 ), then the congruence of (x) b(mod n) is solvable if and only if φ(n) b 2s ≡ 1(mod n). Proof. Let a be a semi-primitive root modulo n and we have b ≡ (a2)k(mod n). At first we suppose that the equation (x2)k ≡ b(mod n) is solvable, then:

φ(n) 2 k φ(n) φ(n) k b 2s ≡ ((x ) ) 2s ≡ (x ) s (mod n) | k Since s k and s is an integer, so by Theorem ”2.6” we have: φ(n) φ(n) k x ≡ 1(mod n) =⇒ b 2s ≡ 1 s ≡ 1(mod n)

φ(n) 2 k Conversely: Let b 2s ≡ 1(mod n) so b ≡ (a ) (mod n) result is that:

φ(n) k( φ(n) ) 1 ≡ b 2s ≡ (a) 2s (mod n)

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8 A. R. ALIZADEH M.

φ(n) Since ordn(a) = 2 than: φ(n) φ(n) |k( ) 2 2s k | 2 k ≡ Therefore 2s is an integer (i.e. 2s k) and hence, the equation of (x ) b(mod n) is solvable. □ Example 2.1. Find all the answers to the following equation x11 ≡ 8(mod 15) Solution: It is obvious that the integer number of 2 is a semi-primitive root of 15 and the number of 15 haven’t any primitive root. So for solving the equation of x11 ≡ 8(mod 15) we have: 24 ≡ 1(mod 15) Let y be any positive integer such that: x ≡ 2y(mod 15), then the equation is to be: 211y ≡ 23(mod 15) But by the Theorem of ”2.6” we have: 211y ≡ 23(mod 15) ⇐⇒ 11y ≡ 3(mod φ(15)) Where: φ(15) = 8. Therefore y ≡ 11−1 × 3(mod 8) ≡ (3)−1 × 3(mod 8) ≡ 1(mod 8) So the final answer is x ≡ 2((mod 15). □ The author suggests that further research in this direction is likely going to reveal additional properties of Semi-primitive lambda-roots and thus contribute to their understanding of how such structures can be defined on the underlying λ roots, where a semi-primitive lambda-root of n is an element of largest possible order(namely, semi(λ(n))) in U(n).

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1. P. Goswami and M. M. Singh, On the number of semi-primitive roots modulo n, Notes on Number Theory and Discrete Math. 21 (2015), No.4, 48-55

2. K. V. Lakshmi, Properties of Semi-primitive roots, IOSR J. Math. 5 (2013), No.4, 08-11.

3. K. Lee, M. Kwon, M. K. Kang and G. Shin, Semi-primitive root modulo n, Honam Math. J. 33 (2011), No.2, 181-186.

4. K. Lee, M. Kwon and G. Shin, Multiplicative groups of integers with semi-primitive roots modulo n, Commum. Korean Math. Soc. 28 (2013), No.1, 71-77.

5. H. E. Rose, A Course in Number Theory, Oxford University Press Inc., New York, (1994).

Author Department of Mathematics, Faculty of Science, University of Yasouj, Yasouj, IRAN E-mail address: [email protected]

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