For arbitrary a and n, let C(a, n) denote the number of cycles in G(a, n), and let c(a, n, d) be the number of cycles in G(a, n) with GCD d.

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1 Directe Graphs Define by Arithmetic (mo n) EZRA BROWN Department of Mathematics Virginia Tech Blacksburg, Virginia , USA 1. Introuction. Let a an n>0beintegers, an efine G(a, n) to be the irecte graph with vertex set V = {0, 1,...,n 1} such that there is an arc from x to y if an only if y ax (mon). Recently, Ehrlich [1] stuie these graphs in the special case a = 2 an n o. He prove that if n is o, then the number of cycles in G(2,n) is o or even accoring as 2 is or is not a quaratic resiue mo n. The aim of this paper is to give the analogous results for all a an all positive n. In particular, we show that if a an n are relatively prime, an n is o, then the number of cycles in G(a, n) is o or even accoring as a is or is not a quaratic resiue mo n. Define GP (a, n) be the irecte graph with vertex set V = {0, 1,...,n 1} such that there is an arc from x to y if an only if y x a (mon). We etermine the number of cycles in GP (a, n) for n a prime power. 2. Preliminary Results. We require a few lemmas. In what follows, write n to mean that is a ivisor of n, an let (x, y) an [x, y] enote the greatest common ivisor (GCD) an least common multiple (LCM), respectively, of x an y. If (a, m) = 1, then (a/m) enotes the familiar Legenre Jacobi quaratic resiue symbol. Finally, let U n = {x :1 x n an (x, n) =1}, let ϕ(n) enote the Euler phi function, an if (a, n) = 1, then let or n (a) be the least positive integer r such that a r 1 (mo n). Lemma 1. Let (a, n) =1.If(x 1,x 2,...,x r ) is a cycle in G(a, n), then (n, x i ) is the same for each i, 1 i r. Proof. Let (x 1,x 2,...,x r ) be a cycle in G(a, n). Since (a, n) = 1, it follows that (n, x 2 )=(n, ax 1 )=(n, x 1 ) = 1, an so for each i, (n, x i )=(n, x 1 ) by inuction. [We shall call this common value of (n, x i ) the GCD of the cycle (x 1,x 2,...,x r ).] For arbitrary a an n, let C(a, n) enote the number of cycles in G(a, n), an let c(a, n, ) be the number of cycles in G(a, n) with GCD. Lemma 2. Let (a, n) =1. Then c(a, n, 1) = ϕ(n) or n (a). For example, let a = 3 an n = 65. Then ϕ(65) = 48, or 5 (3) = 4, or 13 (3) = 3 an so or 65 (3) = 12. Thus, c(3, 65, 1) = 48/12 = 4, an the four relevant cycles are (1, 3, 9, 27, 16, 48, 14, 42, 61, 53, 29, 22), (2, 6, 18, 54, 32, 31, 28, 19, 57, 41, 58, 44), (4, 12, 36, 43, 64, 62, 56, 38, 49, 17, 51, 23), an (7, 21, 63, 59, 47, 11, 33, 34, 37, 46, 8, 24). Proof. Let r = or n (a). Then the elements of the cycle (1,a,...,a r 1 ) form a subgroup <a>of U n of orer r. The claim is that the cosets of <a>in U n an 1

2 2 the cycles in G(a, n) with GCD 1 are in one to one corresponence. For, writing x y to mean that x an y are in the same coset of <a>in U n, we see that x y if an only if x 1 y a i (mo n), for some integer i. But this is precisely the conition that x an y lie on a cycle in G(a, n). Hence, c(a, n, 1) is equal to the number of cosets of <a>in U n, i.e. the inex of <a>in U n. But since the ϕ(n) group U n has orer ϕ(n), this inex is just or n (a). Lemma 3. If (a, n) =1, an n, then c(a, n, ) =c(a, n, 1). For example, the cycles in G(2, 45) with GCD 3 are (3, 6, 12, 24) an (21, 42, 39, 33); the corresponing cycles in G(2, 15) with GCD 1 are (1, 2, 4, 8) an (7, 14, 13, 11). Proof. Let (x 1,x 2,...,x r ) be a cycle in G(a, n) with GCD. Then x 2 ax 1,..., x r a r 1 x 1 an x 1 a r x 1 (mo n) with r positive an minimal. This is true if an only if (1,a,...,a r 1 n ) is a cycle in G(a, (n, x 1 ) )=G(a, n ) (clearly with GCD 1). Hence, each cycle in G(a, n) with GCD has length r = or n/ (a). Furthermore, x an y lie on a cycle in G(a, n) with GCD if an only if y xa i (mo n), i.e. y x ai (mo n ) which is precisely the conition that x an y lie on a cycle in G(a, n ). Hence the number of cycles in G(a, n) with GCD is the same as the number of cycles in G(a, n ) with GCD 1. That is, c(a, n, ) =c(a, n, 1). We are now reay for the main result of this section. THEOREM A. If (a, n) =1, then C(a, n) = n ϕ() or (a). Thus, Proof. For, C(5, 77) = ϕ(1) or 1 (5) + ϕ(7) or 7 (5) + ϕ(11) or 11 (5) + ϕ(77) or 77 (5) = = =6. C(a, n) = c(a, n, ) n = n c(a, n, 1) (by Lemma 3) = c(a,, 1) n = n ϕ() or (a) (by reorering the sum) (by Lemma 2).

3 3. The parity of C(a, n) for (a, n) =1. Next, we etermine the parity of the number of cycles in G(a, n) with GCD 1; from that, we etermine the parity of C(a, n) for (a, n) =1. Lemma 4. Let p be an o prime, let r be a positive integer an let (a, p) =1. Put p 1=2 s q, where q is o. (a) If (a/p) =1, then or p r(a) 2 s 1 qp r 1. (b) If (a/p) = 1, then 2 s or p r(a). Proof. Euler s criterion for the Legenre symbol states that (a/p) a (p 1)/2 (mo p). Thus, if p 1=2 s q, where q is o, then (a/p) a 2s 1q (mo p). We have two cases: (a) If (a/p) = 1, then a 2s 1q 1 (mo p), so that or p (a) 2 s 1 q. If the statement is true for some r 1, then a 2s 1 qp r 1 = 1 + kp r ; raising both sies to the pth power, we have a 2s 1 qp r =(1+kp r ) p 1 (mo p r+1 ). Hence, or p r(a) 2 s 1 qp r 1 by inuction. (b) If (a/p) = 1, then a 2s 1q 1 (mo p), so that 2 s or p (a). Since or p (a) is a ivisor of or p r(a) for r 1, we are one. Lemma 5. Let (a, n) = 1 with n o. If n = p r, where p is a prime an if (a/p) = 1, then c(a, n, 1) is o; in all other cases, c(a, n, 1) is even. Proof. Let p 1 =2 s q, where q is o. By Lemma 4, if (a/p) = 1, then or p r(a) = 2 s k with k o. Since ϕ(p r )=p r 1 (p 1) = p r 1 2 s q, it follows from Lemma 2 that c(a, p r, 1) = ϕ(pr ) or p r(a) = pr 1 q, k which is an o number. Hence c(a, p r, 1) is o. We must now show that in all other cases, c(a, n, 1) is even. First, if n = p r with p as above, an if (a/p) = 1, then the highest power of 2 iviing or p r(a) is2 s 1. Since 2 s ϕ(p r ), it follows that the fraction ϕ(pr ) or p r (a) is even. Next, if n = g i=1 pei i with g>1 an p i 1=2 si q i, then or n (a) [ p e s1 q 1,...,pg eg 1 2 sg q g ] = g i=1 p ei 1 [ ] i q1,...,q g 2 M, where M = max(s 1,...,s g ). Now let S = g i=1 s i. Since n is ivisible by at least two istinct o primes, it follows that S > M, so that c(a, n, 1) = ϕ(n) or n (a) is ivisible by 2 S M. Hence, c(a, n, 1) is even. A slight moification of the above proof yiels the following: Lemma 6. Let (a, n) =1with n even. (a) If n is ivisible either by 8 or by more than one o prime, or if n =4p e with p an o prime, then c(a, n, 1) is even. (b) If p is an o prime, then c(a, p e, 1) = c(a, 2p e, 1). (c) c(a, 1, 1) = c(a, 2, 1) = 1 an c(a, 4, 1) = ( 1/a)+3. 2 We may now prove our main results. 3

4 4 THEOREM B. Let a an n be relatively prime, an let n be o. Then the number of cycles in G(a, n) is o or even accoring as a is or is not a quaratic resiue mo n. That is, C(a, n) 1+(a/n) ( mo 2). 2 For example, C(3, 1001) is even because (3/1001) = (1001/3) = (2/3) = 1. A bit of irect calculation reveals that or 7 (3) = 6,or 11 (3) = 5 an or 13 (3) = 3, so that C(3, 1001) = ϕ() or (a) 1001 = = =54, which is inee even. Somewhat more tricky is the evaluation of C(2159,pq), where p = an q = are both primes. However, since pq 3 (mo 4), we see that (2159/pq) = (pq/2159) = (743/2159) = (2159/743), which reuces to the prouct (2/673)(8/35), or 1. Hence C(2159,pq) is even. Proof. Let n = g i=1 pei i with each p i o, an suppose (a, n) = 1. It follows from Theorem A an Lemma 5 that C(a, n) = n ϕ() g or (a) 1+ e i i=1 j=1 ϕ(p j i ) or p j (a) (mo2), i since all other terms are even. If we orer the primes p i so that for some integer f (which might be 0), (a/p i ) = 1 if an only if i>f, then we see that e i C(a, n) 1+ 1 (mo 2) i f j=1 1+ e i (mo 2). i f On the other han, since n is o an (a, n) = 1, we use the well known properties of the Legenre an Jacobi symbols to see that (a/n) = g (a/p i ) ei i=1 = i f( 1) ei (since (a/p i )=1fori>f) =( 1) i f ei, so that ( 1) C(a,n) ( 1) 1+ i f ei (a/n) (mo 2). Hence C(a, n) is o if (a/n) = 1, an C(a, n) isevenif(a/n) = 1, an we are one.

5 5 THEOREM C. Let a an n be relatively prime, let n be even, an write n =2 e n, where n is o. (a) If e =1, then G(a, n) has an even number of cycles. (b) If e 2, then the number of cycles in G(a, n) is even or o accoring as 1 is or is not a quaratic resiue mo n. That is, C(a, n) 1 ( 1/n ) ( mo 2). 2 Proof. Theorem C follows from Theorem A an Lemma 6 in the same way thattheorem B follows from Theorem A an Lemma The parity of C(a, n) for arbitrary a an n. We are now reay to exten Theorems B an C to the graphs G(a, n), where a an n are not relatively prime. The principal observation is the corresponence between the cycles in G(a, qm) an the cycles in G(a, m). Specifically, we have the following: Lemma 7. Suppose (m, a) =1an suppose that each prime ivisor of q ivies a. Then C(a, qm) =C(a, m). Proof. Let x be an integer mo qm. We may write x =(x a,y), where (y, a) =1 an each prime ivisor of x a ivies a. Thus, (x a,q)=1. Nowleti 0 an r>0 be minimal an satisfy a i+r x a i x (mo qm). This happens if an only if y(a r 1)(a i x a ) 0 (mo qm). But (a i x a,q) = 1, an (y(a r 1),m) = 1. Hence, the above congruence hols if an only if q y(a r 1) an m a i x a. Thus, (a i x,...,a i+r 1 x) is a cycle in G(a, qm) if an only if i is the least nonnegative integer such that m a i x an (y,ay,...,a r 1 y) is a cycle in G(a, q), where y is the largest ivisor of x relatively prime to m. But this means that the cycles of G(a, qm) an the cycles of G(a, q) are in one to one corresponence, i.e. C(a, qm) =C(a, m). As a irect consequence of Lemma 7, we have the following result: THEOREM D. If a an n are positive integers, then the parity of C(a, n) is equal to the parity of C(a, n ), where n is the largest ivisor of n that is relatively prime to a. 5. The cycle structure of the graphs GP (a, n) for n a prime. Let GP (a, n) be the irecte graph with vertex set V = {0, 1,...,n 1} such that there is an arc from x to y if an only if y x a (mon). Let CP(a, n) enote the number of cycles in the graph GP (a, n). There are some interesting ifferences between the graphs GP (a, n) an G(a, n). For example, if (a, n) = 1, then every vertex of G(a, n) lies on a cycle. This is not the case for the vertices of GP (a, n). If p n is a prime power, then GP (a, p n ) looks like a union of charm bracelets, with each charm a tree that correspons to a coset of a certain subgroup U of roots of unity mo p n. In particular, if we write ϕ(p n )=qr, where (q, a) = 1, every prime ivisor of r ivies a, an m is the least positive integer such that r a m, then U consists of the a m th roots of unity mo ϕ(p n ). Our principal result of this section is the following theorem:

6 6 THEOREM P. If p n is an o prime, then there is a one to one corresponence between the cycles of GP (a, p n ) an the cycles of G(a, q), where q is the largest ivisor of ϕ(p n ) that is relatively prime to a. Furthermore, CP(a, p n )=1+ ϕ(p n ), (,a)=1 ϕ() or (a). The following lemma leas us to the proof of Theorem P. Lemma 8. Let p n be a prime power; let g be a primitive root (mo p); let (a, p) =1 an write ϕ(p n )=qr, where (q, a) =1an every prime ivisor of r ivies a. Then x an y lie on a cycle in GP (a, p n ) if an only if either (a) there exist integers j an k such that x g rj (mo p n ), y g rk (mo p), an j an k lie on a cycle of G(a, q), or (b) x = y =0. Proof. If p x, then for some positive integer s, x as 0 (mo p n ). Thus, if p x, then x lies on a cycle in GP (a, p n ) if an only if x 0 (mo p n ). From here on, we assume that x an y are relatively prime to p. If x is a vertex of GP (a, p n ), then we may write x g t (mo p n ) for some integer t with 0 t<ϕ(p n ). Let us first show that x lies on a cycle of GP (a, p n ) if an only if r t. We have the following sequence of equivalent statements: x lies on a cycle of GP (a, p n ) if an only if x as x (mop n ) for some positive integer s if an only if g t(as 1) 1 ( mo p n ) for some positive integer s if an only if ϕ(p n ) t(a s 1). Hence, if x lies on a cycle of GP (a, p n ), then rq t(a s 1). Now each prime ivisor of r ivies a, so it follows that (r, a s 1) = 1. We conclue that r t. Conversely, suppose that r t, so that x g rj (mo p n ) for some integer j. If j = 0, then x = 1, which is clearly on its own cycle; since g ϕ(pn) 1 (mo p n ), we may assume that 1 j q 1. The above argument shows that x is on a cycle if an only if rq rj(a s 1) for some integer s. Since 1 j q 1, it follows that q (a s 1). In particular, if s = or q (a), then we may conclue that x lies on a cycle of length s. Next, x an y will lie on a common cycle if an only if x g rj (mo p n ) an y g rk (mo p n ) lie on a common cycle of GP (a, p n ). It is straightforwar to verify that this happens if an only if there exists an integer m such that ja m k (mo q) i.e., that j an k lie on a cycle of G(a, q). Finally, if (j,ja,...,k ja m,...,ja s 1 ) is a cycle in G(a, q), then it follows that s = or q (a), which means that is a cycle in PG(a, p n ), an we are one. (g rj,g rja,...,g rjam,...,g rjas 1 ) Theorem P now follows from Lemma 8 an Theorem A, an from the fact that there is one extra cycle in PG(a, p n ) the cycle consisting of the irecte loop from the vertex 0 to itself.

7 7 REFERENCE [1.] Amos Ehrlich, Cycles in oubling iagrams mo m, Fibonacci Quarterly 32 (1994), AMS Classification Numbers: 11A07, 05D20, 11A15