PERIODS OF FIBONACCI SEQUENCES MODULO m. 1. Preliminaries Definition 1. A generalized Fibonacci sequence is an infinite complex sequence (g n ) n Z

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1 PERIODS OF FIBONACCI SEQUENCES MODULO m ARUDRA BURRA Abstract. We show that the Fiboacci sequece modulo m eriodic for all m, ad study the eriod i terms of the modulus.. Prelimiaries Defiitio. A geeralized Fiboacci sequece is a ifiite comlex sequece (g ) Z which satisfies the recurrece g = g + g for all. The Fiboacci sequece is the geeralized Fiboacci sequece (F ) for which F 0 =0ad F =. If (g ) is ay geeralized Fiboacci sequece, ad m Z, the we ca costruct a ew sequece (g (m) ) such that for each, g (m) is the remaider of g whe divided by m. The sequece (F (m) ) is called the sequece of Fiboacci umbers modulo m. Lemma. If (g ) is a geeralized Fiboacci sequece i Z, g (m) is a geeralized Fiboacci sequece i Z/(m). Proof. The roof follows from the fact that the caoical rig homomorhism from Z to Z/(m) seds ay g Z to its remaider whe divided by m, which is g (m). Defiitio. If (a ) Z is a ifiite comlex sequece, we defie the eriod of (a ) to be the least umber such that (a + )=(a ). We rovide the followig lemma without roof. Lemma. For ay eriodic sequece (a ), a + = a for all if ad oly if the eriod of (a ) divides. Theorem. The sequece (F (m) ) is eriodic for all m. The eriod of this sequece is if ad oly if is the least umber such that F (m) =0ad F (m) + =. Proof. Recall that geeralized Fiboacci sequeces which agree o cosecutive values are idetical: i other words, if (g ) ad (h ) are geeralized Fiboacci sequeces, ad g = h, g + = h + for some, the (g )=(h ). We claim (without roof) that Lemma shows that this is also true for geeralized Fiboacci sequeces modulo m. ( Now for ) ay m, F (m) ca oly tae fiitely may values, ad hece the air F (m),f (m) + ca oly tae fiitely may values. Hece oe of these airs must reeat; i.e., F (m) 0 = F (m) (m) 0+ ad F = F (m) 0+ for some 0++ 0,. Thus F (m) = F (m) +, for all, so F (m) is eriodic. We deote its eriod by P (m). Date: May 000.

2 ARUDRA BURRA Now suose is the least umber such that F (m) =0=F (m) 0 ad F (m) + == F (m). The the sequeces F (m) ad F (m) + agree o cosecutive values, ad hece are idetical. Suose P (m) =j for some j<(by Lemma,j, soj ). The, i articular, F (m) j = F (m) 0 = 0 ad F (m) j+ = F (m) =, which cotradicts the choice of as the least umber for which this was ossible. Hece = P (m). Coversely, suose P (m) =. The F (m) = F (m) 0 = 0 ad F (m) + = F (m) =. If this also holds for some j<, the it follows from we have just roved that j = P (m), which is a cotradictio. Hece must be the least umber for which F (m) = 0 ad F (m) + =. The followig corollary is a easy cosequece of Lemma ad Theorem : Corollary. For ay m, F 0 (mod m) ad F + (mod m) if ad oly if P (m).. Periods of F (m) for rime moduli Theorem. Let > be rime. If ± (mod ), the P (). If ± (mod ) the P () +. We will require the followig lemma: Lemma 3. F = [ Proof. By the Biet formula for F, ( ( mod ) ) +(mod) [( F = + ) ( ) ] [ + = ( + + +( ) + +( ) )] = [ ( ) ] (mod) 3 +( mod ) ] Theorem follows from the followig fact, which we rove usig Lemma 3 ad some elemetary results i umber theory. If is rime, recall that (the Legedre symbol) is whe a is a quadratic residue modulo, ad whe it is ot. Theorem 3. If > is rime, the () F (mod ) a () F ( 0 (mod ) )

3 PERIODS OF FIBONACCI SEQUENCES MODULO m 3 Proofof. Sice >is rime, it must be odd, hece +(mod ) = +. Lemma 3 gives us the followig exressio for F : F = [ ] Now observe that (mod ) by Fermat s Little Theorem, ad divides the biomial coefficiets (,..., ).Sowehave F ] [ (mod ). (mod ). By Euler s criterio for quadratic residues, (mod ). Hece F (mod ). Proofof. Suose = ; i.e., is a quadratic residue modulo. The a (mod ) for some a. Notice that the exasio of F i 3 does ot have ay terms cotaiig. Thus we may substitute a for i the Biet formula for F,to get ( F + ) ( ) (mod ) [ ( + a) a ( a) ] (mod ). Alyig Fermat s Little Theorem, we see that (+a) ( a) (mod ), so F ( ) F 0 (mod ). Now suose =. Sice is odd, + is eve, ad we ca use Lemma 3 to write F + = [ ] Now ote that ( + = + ) ( = + (mod ), ad divides each of + ) ( 3,..., + So F + [ ] + (mod ). By Euler s criterio, =, so F ( ) F + 0 (mod ). We ow rove Theorem. Proof. Suose ± (mod ). Sice ad are quadratic residues modulo, ( ) =. By quadratic recirocity (sice is of the form 4+), = =. By Theorem 3, F 0 (mod ), ad F (mod ). By Corollary, P (). Now suose ± (mod ). The is a quadratic o-residue mod, so is a quadratic o-residue mod. Hece =. By Theorem 3, F (mod ) ad F + 0 (mod ). Now F = F L for all, sof + = F (+) = F + L +. Sice F + 0 (mod ), F + 0 L + 0 (mod ). Furthermore, F +3 = F (+)+(+) = F + F + + F + F +. Sice F + 0 (mod ), it follows ).

4 4 ARUDRA BURRA that F +3 F + F + (mod ). By the Fiboacci recurrece, F + = F + F +, so F + (mod ). Thus F +3 ( ) (mod ). By Corollary, P () +. Fially, for the cases = ad = we may verify by calculatio that the eriod of (F () ) is 3, ad the eriod of (F () )is0. 3. Periods of F (m) for owers of rime moduli Theorem 4. Let > be rime. For ay r, P ( r ) P () r. We will require the followig lemma: Lemma 4. For m>, P (m) is eve. Proof. Let = P (m), ad suose is odd. Recall that F F F + =( ) for ay. I articular for =, we have F F F + =, ad hece F F F + (mod m). By Corollary, F 0 (mod m) ad F F + (mod m). Thus F F F + (mod m). Hece (mod m), which is ot ossible for m>. Proof of Theorem 4. We roceed by iductio. First cosider the case r =, ad let = P (). Recall the followig idetity, which holds whe m is odd: (3) F m F = L (m ) +( ) L (m 3) + L (m ) + +( ) m 3 L +( ) m Sice is a rime, >, is odd; by Lemma 4, = P () is eve. Substitutig ad for m ad resectively i 3, we get (4) F F = L ( ) + L ( 3) + + L + Now L ( i) = F ( i) + F ( i)+ for ay i. Sice = P (), F ( i) F (mod ) ad F ( i)+ F (mod ). Now F = F =,sol ( i) (mod ) for i =, 3,...,. There are such terms i 4. Thus F F + 0 (mod ). Hece F F,soF F. But = P (), so F 0 (mod ); i.e., F.Thus F ; i.e., F 0 (mod ). We ow show that F + (mod ). Sice = P () is eve, F F F + =( ) =. Sice F 0 (mod ), F F + (mod ). But F F + (mod ) by the Fiboacci recurrece, so F+ (mod ), therefore (F + ) (F + + ). There are three ossibilities:. F + ad F + +. F F + Suose is the case. The (F + +) (F + ) =, which is ot ossible sice >. Now cosider case. If F + + the F + +; i.e., F + (mod ). But by Corollary, F + (mod ) sice = P (). This is a cotradictio sice (mod ) caot hold for >. This leaves oly the ossibility that F +. Therefore F + (mod ). We have show that F 0 (mod ) ad F + (mod ). From Corollary it follows that P ( ) P ().

5 PERIODS OF FIBONACCI SEQUENCES MODULO m The argumet used to show that F r 0 (mod r ) for r = geeralizes immediately for arbitrary r, sice it deeds oly o the (uchaged) facts that is eve ad r is odd. Similarly, the roof that F r + (mod r ) for r = ca be alied to the case of arbitrary r, which allows us to comlete the iductio ste. 4. Periods of F (m) for roducts of relatively rime moduli We coclude with the followig theorem, which relates the eriod of (F (q) the eriods of (F () ) ad (F (q) ) whe (, q) =. Theorem. Let (, q) =. The P (q) = lcm(p (),P(q)). Proof. Let = P (q). By Corollary, F 0 (mod q) ad F + (mod q), so q F, ad q F +. Hece F ad F + ; i.e., F 0 (mod ) ad F + (mod ). By Corollary, P (). Similarly, P (q). Hece lcm(p (),P(q)) P (q). Coversely, let a = P (), b = P (q). Sice lcm(a, b) is a multile of both a ad b, F lcm(a,b) 0 (mod ) ad F lcm(a,b) 0 (mod q). Sice (, q) =, by the Chiese Remaider Theorem, F lcm(a,b) 0 (mod q). Similarly, F lcm(a,b)+ (mod q). So lcm(a, b); i.e. P (q) lcm(p (),P(q)). Hece P (q) = lcm(p (),P(q)).. Periods of Geeralized Fiboacci Sequeces modulo m ( Theorem ) 6. Let (g ) be ay geeralized Fiboacci sequece i Z. For ay m, g (m) is eriodic. If P g (m) deotes the eriod of this sequece modulo m, the P g (m) P (m). Proof. Recall that {(F ), (F + )} is a basis for the sace of geeralized Fiboacci sequeces over Z. Hece ay geeralized Fiboacci sequece (g ) ca be exressed as a liear combiatio of these two sequeces. Let g = af + bf + for some a ad b. Substitutig 0 ad for F 0 ad F,wehaveg 0 = b ad g = a + b, whece get a = g g 0 ad b = g 0.Thusg =(g g 0 )F + g 0 F + for all. If = P (m), g (m) + =(g(m) g (m) 0 )F (m) + + g(m) 0 F (m) ++ =(g(m) g (m) 0 )F (m) + g (m) 0 F (m) + = g(m) for ay (these calculatios are licesed by the caoical rig homomorhism metioed i Lemma ). Hece (g ) is eriodic, ad, by Lemma, P g (m) P (m) for ay m. Theorem 7. Let (g ) be ay geeralized Fiboacci sequece i Z, ad let be g0 g rime. If ( g g ) is ivertible i Z/() the P g () =P (). Proof. We claim that F = g0g+ gg. To verify this, we eed to chec oly the g0 +g0g g cases = 0 ad =. For = 0, the umerator is g 0 g g g 0 =0=F 0. For =, the umerator is g 0 g g = g 0 (g 0 + g ) g = g0 + g 0 g g which is equal g0 g to the deomiator, so the fractio evaluates to = F. If ( g g ) is ivertible i g0 g Z/(), the ) ca be exressed as a liear combiatio of (g () claim that P () P g () P g () =P (). )to g g = g0 +g 0 g g ad hece (F () ) ad (g () + ), ad we ca aly the argumet of Theorem 6 to. Combiig this with the result of Theorem 6, we see that