Table Of ContentThe largest cycles consist by the
6
quadratic residues and Fermat primes
1
0
2
Haifeng Xu˚
n
a Yangzhou University
J
7 January 29, 2016
2
]
T
Abstract
N
Thispaperstudiesthelargestcyclesconsistedbythequadraticresidues
.
h modulo prime numbers. We give some formulae about the maximum
at length of thecycles. Especially, theformula for moduloFermat primes is
m given.
[ MSC2010: 11A07.
Keywords: quadratic residues, Fermat primes, largest cycles.
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v
9
1 Examples and Definition
0
5
6 Wefindthisphenomenonbycomputation. Soletusbeginwithsomeexamples.
0
Let’s consider quadratic residues module 999 first.
.
1
0 4542 ”322 pmod 999q, 4452”223 pmod 999q,
6
3222 ”787 pmod 999q, 2232”778 pmod 999q,
1 $ $
Xiv: ’’’’’’’’&79887822 ””918281 ppmmoodd 999999qq,, ’’’’’’’’&78788922””818192 ppmmoodd 999999qq,,
1212 ”655 pmod 999q, 1122”556 pmod 999q,
r
a Wefindth’’’’’’’’%at65th52es”en4u5m4bpmerosdt495949,q3,22,787,’’’’’’’’%98585,6122”1,645455upcmoondsis9t9a9qc.ycle,alsofor
the numbers t445,223,778,889,112,556u. The length of the cycle is 6. In fact
itisthemaximumnumberoftheelementsinthecycles. We calleditthelength
of the largest cycles for the quadratic residues of 999. Look another example
˚Project supported by NSFC(Grant No. 11401515), the University Science Research
Project of Jiangsu Province (14KJB110027) and the Foundation of Yangzhou University
2014CXJ004.
1
for modulus 99, we get
702 ”49 pmod 99q,
222 ”88 pmod 99q, 492 ”25 pmod 99q,
$
#882 ”22 pmod 99q, ’’’&252 ”31 pmod 99q,
312 ”70 pmod 99q.
Weseethat,thereexistasmallercycle. W’’’%eareinterestinginthelargestcycles.
Definition. Consider the equation x2 ” a pmod mq. If there exist a series of
numbers tx uk such that
i i“1
x2 ”x pmod mq,
1 2
$x22 ”x3 pmod mq,
’’’’’’&x2k´1 ”¨¨¨xk pmod mq,
x2 ”x pmod mq,
thenwecalltheseknumber’’’’’’%scoknsistacy1clemodulom. Thenumberkisdefined
as the length of the cycle.
It infers that
x2k ”x pmod mq, for i“1,2,...,k.
i i
Thereexists atleastonelargestcycle. We denote the lengthofit by Lpmq. For
example, Lp99q “ 4, Lp999q “ 6. In the next section, we try to find out the
formula for Lpmq.
2 Computations
The maximum length of the cycles modulo prime numbers are list as follows:
m“p 2 3 5 7 11 13 17 19 23 29
Lppq 1 1 1 2 4 2 1 6 10 3
m“p 31 37 41 43 47 53 59 61 67 71
Lppq 4 6 4 6 11 12 28 4 10 12
m“p 73 79 83 89 97 101 103 107 109 113
Lppq 6 12 20 10 2 20 8 52 18 3
m“p 127 131 137 139 149 151 157 163 167 ...
Lppq 6 12 8 22 36 20 12 54 82 ...
Obviously, it establishes a map from the set of primes PRIMES to the set
of positive numbers Z`:
L: PRIMES Ñ Z`,
p ÞÑ Lppq.
2
Infact,E.L.Blanton,Jr.,S.P.HurdandJ.S.McCranie[8]hadconsidered
this question. We can find this sequence on the site of OEIS [4].
For composites, we have
m“c 4 6 8 9 10 12 14 15 16 18 20 21
Lpcq 1 1 1 2 1 1 2 1 1 2 1 2
m“c 22 24 25 26 27 28 30 32 33 34 35 36
Lpcq 4 1 4 2 6 2 1 1 4 1 2 2
m“c 38 39 40 42 44 45 46 48 49 50 51 ...
Lpcq 6 2 1 2 4 2 10 1 6 4 1 ...
And when modulo p2, we have
m“p2 22 32 52 72 112 132 172 192 232 292
Lpp2q 1 2 4 6 20 12 8 18 110 84
m“p2 312 372 412 432 472 532 592 612 672 712
Lpp2q 20 36 20 42 253 156 812 60 330 420
m“p2 732 792 832 892 972 1012 1032 1072 1092 1132
Lpp2q 18 156 820 110 48 100 408 2756 36 84
m“p2 1272 1312 1372 1392 1492 1512 1572 1632 1672 ...
Lpp2q 42 780 136 1518 1332 60 156 162 6806 ...
And consider modulo the powers of prime numbers. For examples:
3
m“pn 4 8 16 32 64 ...
Lppnq 1 1 1 1 1 ...
m“pn 32 33 34 35 36 ...
Lppnq 2 2ˆ3 3ˆ6 6ˆ9 9ˆ18 ...
m“pn 52 53 54 55 56 ...
Lppnq 4 4ˆ5 5ˆ20 20ˆ25 25ˆ100 ...
m“pn 72 73 74 75 76 ...
Lppnq 6 6ˆ7 7ˆ42 42ˆ49 ... ...
m“pn 112 113 114 115 116 ...
Lppnq 20 20ˆ11 220ˆ11 26620 ... ...
m“pn 132 133 134 135 136 ...
Lppnq 12 12ˆ13 13ˆ156 ... ... ...
m“pn 172 173 174 175 176 ...
Lppnq 8 8ˆ17 17ˆ136 ... ... ...
m“pn 192 193 194 195 196 ...
Lppnq 18 18ˆ19 342ˆ19 ... ... ...
m“pn 232 233 234 235 236 ...
Lppnq 110 110ˆ23 ... ... ... ...
m“pn 292 293 294 295 296 ...
Lppnq 84 84ˆ29 ... ... ... ...
m“pn 312 313 314 315 316 ...
Lppnq 20 20ˆ31 ... ... ... ...
m“pn 472 473 474 475 476 ...
Lppnq 253 253ˆ47 ... ... ... ...
The data listed in these tables are all verified by computations.
Lpp2q is relevant to p´1 and Lppq. It is rather complicated and we discuss
it later. But for few exceptions we have the explicit formula. For example, the
Fermat primes (Proposition 4.1 and 4.3).
For the first five Fermat primes F “ 3, F “ 5, F “ 17, F “ 257,
0 1 2 3
F “65537,we have LpF q“1, i“0,1,2,3,4.
4 i
For p2 “112, we have Lp121q“20. The largest cycle tx ,x ,...,x u is
1 2 20
4,16,14,75,59,93,58,97,92,115,36,86,15,104,47,31,114,49,102,119.
Of course, there are also cycles with length equal to 10. For example, one of
them is
100,78,34,67,12,23,45,89,56,111.
For p2 “ 172, we have Lp289q “ 8 “ 1p17´1q. For example, one of the
2
cycles is
256,222,154,18,35,69,137,273.
For p2 “ 2572, we have Lp66049q“ 16 “ 1 p257´1q. For example, one of
24
4
cycles is
65536, 65022, 63994, 61938, 57826, 49602, 33154, 258,
515, 1029, 2057, 4113, 8225, 16449, 32897, 65793.
For p2 “ 655372, we have Lp4295098369q “ 32 “ 1 p65537 ´ 1q. For
211
example, one of the cycles is
4294967296, 4294836222, 4294574074, 4294049778,
4293001186, 4290904002, 4286709634, 4278320898,
4261543426, 4227988482, 4160878594, 4026658818,
3758219266, 3221340162, 2147581954, 65538,
131075, 262149, 524297, 1048593,
2097185, 4194369, 8388737, 16777473,
33554945, 67109889, 134219777, 268439553,
536879105, 1073758209, 2147516417, 4295032833
Note that
17“222 `1, 257“223 `1, 65537“224 `1.
They are the Fermat numbers. For k “0,1,2,3,4,22k `1“3,5,17,257,65537
are all prime numbers. But 232 `1 “ 4294967297 is not a prime. It equals
641ˆ6700417. It is conjectured that there are only 5 terms. Currently it has
been shown that 22k `1 is composite for 5ďk ď32 [2], [3].
3 Some lemmas
Definition ([7]). Let n P N. A primitive root mod n is a residue class α P
pZ{nZq˚ with maximal order, i.e., ordpαq“ϕpnq.
Lemma 3.1. The Euler’s totient function ϕpmq have the following formula for
m“pn:
ϕppnq“pp´1qpn´1.
Euler’s totient function is a multiplicative function, meaning that if two
numbers m and n are coprime, then ϕpmnq“ϕpmqϕpnq.
Lemma 3.2 (Euler’s Criterion). Let p be an odd prime and a not divisible by
p. Then a is a quadratic residue modulo p if and only if
app´1q{2 ”1 pmod pq.
The proof uses the fact that the residue classes modulo a prime number are
a field.
Lemma 3.3 ([5], [6]). The set of all quadratic nonresidues of a Fermat prime
is equal to the set of all its primitive roots.
5
Proof. Let a be a quadratic nonresidue of the Fermat prime F , and let e “
n
o1rd“Fn2a2n..AIctcofrodllionwgstothFaetrmea“t’s2klitftoler tshoemoerenmo,nanFeng´at1iv”e1kpďmo2dn.FnOq.nStohee|oFtnhe´r
hand, by Euler’s criterion,
apFn´1q{2 “222n´1 ”´1 pmod F q.
n
Hence, if k ă 2n, then 2k|22n´1 and so a22n´1 ” 1 pmod F q, which is a
n
contradiction. So, k “ 2n and ord a “ 22n. Therefore, a is a primitive root
Fn
modulo F .
n
Lemma 3.4 ([7]). Let nPN, aPZ coprime to n. Consider the equation
xd ”a pmod nq
paq It has a solution if and only if aϕpnq{gcdpd,ϕpnqq ”1 pmod nq.
pbq If it has a solution, it has exactly gcdpd,ϕpnqq solutions modulo n.
pcq If x is a solution to this equation, then any other solution x1 satisfies x1 ”
yx pmod nq for a unique solution y mod n to yd ”1 pmod nq.
On one extreme, this means if gcdpd,ϕpnqq“1, then there is a unique d-th
rootmodnofanyaPZcoprimeton. Ontheotherhand,itmeansthatforany
d|ϕpnq, there are exactly d solutions mod n to the equation xd ” 1 pmod nq.
This in fact characterizes the existence of primitive roots:
Lemma 3.5. Let nPN. The following are equivalent:
paq There is a primitive root mod n.
pbqForalldPNwithd|ϕpnq, thereareexactlydsolutionsmodntotheequation
xd ”1 pmod nq.
pcq For all d P N with d|ϕpnq, there are exactly ϕpdq elements of pZ{nZq˚ of
order d.
Corollory3.6. Givenanyoddprimenumberp. Letd“gcd 2p´1´1,pp´1qp .
Then there are exactly d solutions mod p2 to the equation
` ˘
xd ”1 pmod p2q.
4 Main results
Proposition 4.1. For the Fermat primes, i.e., the prime numbers of the form
p“22k `1, we have Lppq“1.
Proof. Suppose F “22k `1 is a prime, k ě1. Then ϕpF q“22k. To find the
k k
largest cycle of the quadratic residue equation of modulo F . We consider the
k
equation
x2m ”x pmod F q,
k
6
where x is coprime with F . Then it is equivalent to the equation
k
x2m´1 ”1 pmod F q. (.)
k
By Lemma 3.4, (.) have a solution. And the number of the solutions is equal
to
gcdp2m´1,22kq“1.
Since x “ 1 is the trivial solution of (.), hence there is no other solutions.
Hence we have LpF q“1.
k
Remark 4.2. Since there is a primitive root mod F , by Lemma 3.5, for any
k
1ďm ď2k, x2m ”1 pmod F q have exact 2m solutions. Suppose k ě1. Let’s
k
consider the equation
x2 ”´1 pmod F q.
k
By Lemma 3.4, it has a solution since p´1q22k´1 ” 1 pmod F q. And there
k
are exactly two solutions of this equation. By Lemma 3.3, these two solutions
tx ,x uaretheprimitiverootsmoduloF . Bydefinitionofprimitiveroot,they
1 2 k
havethemaximalorder. Thatis,ordpx q“ordpx q“ϕpF q“22k. Hence,any
1 2 k
othernumbera coprimetoF (notthe primitiveroot)withorderordpaqă22k.
k
And ordpaq|22k´1. Hence, we have
a22k´1 ”1 pmod F q. (.)
k
It infers that the equationx2 ”a pmod F q has a solutionfor any a coprime to
k
F . (Use Lemma 3.4 again.)
k
Proposition 4.3. For the Fermat primes, i.e., the prime numbers of the form
p“22k `1, we have
1
Lpp2q“Lpp22k `1q2q“ pp´1q“2k`1.
22k´k´1
Proof. We have verified it for k “ 0,1,2,3,4, the known Fermat primes. Sup-
pose p “ F “ 22k `1 is a prime, k ě 1. Similar to the Proposition 4.1, we
k
consider the equation
x2m ”x pmod F2q, x‰F ,
k k
and find the nontrivial solution for the maximal m. Since F is a prime, for
k
x‰F , x is coprime with F2. Then it is equivalent to
k k
x2m´1 ”1 pmod F2q. (.)
k
ByLemma3.4,therearegcdp2m´1,ϕpF2qqsolutions. ByLemma3.1,ϕpF2q“
k k
pF ´1qF “22kp22k`1q. Thus, if gcdp2m´1,22k`1qą1, the equation(.)
k k
has nontrivial solutions. The number of the solutions is equal to 22k `1.
7
Because F “22k `1 is a prime number,
k
gcd 2m´1,22k `1 ą1 ô p22k `1q p2m´1q.
It infers that m“`h¨2k`1, here h˘ě1. And there areˇˇ22k `1 solutions include
the trivial x“1. So, for example, if h“2, then
x2m “x22k`1¨22k`1 “ x22k`1 22k`1 ”x22k`1 ”x pmod F2q.
k
´ ¯
Hence, LpF2q“2k`1 for Fermat prime F .
k k
Sincetherearepp´1q{2quadraticresiduesmodp,wehaveLppqďpp´1q{2.
However,1 mapsto itselfunder squaringmodp,sowe expectLppqďpp´3q{2.
In the theorem below, we state a condition for this to happen.
Theorem 4.4. Let p be a non-Fermat prime. Lppq “ pp´3q{2 if and only if
q “pp´1q{2 is prime and 2 is a generator of pZ{qZq˚.
Proof. Letdbeanodddivisorofp´1. Thenthereisasolutiontotheequations
xd ”1 pmod pq, xi ı1 pmod pq, (.)
1ďiăd. Let x satisfy (.). Then such x is a solution to the equation
x2m ”x pmod pq
forallmsuchthatord p2q|mwhereord p2qisthemultiplicativeorderof2mod
d d
d. Furthermore,
Lppq“ max ord p2qď max d´1.
d
d|p´1 d|p´1
d”1mod2 d”1mod2
We need only consider the case where pp´1q{2 is odd, since if d ă pp´1q{2,
then ord p2q ă pp´3q{2. Now any odd d dividing pp´1q satisfies ord p2q ď
d d
ord p2q, hence Lppq “ ord p2q. This is pp´3q{2 if and only if q “
pp´1q{2 pp´1q{2
pp´1q{2 is prime and 2 is a generator of pZ{qZq˚.
Remark 4.5. Similar to the case for Lppq, we have
Lpp2q“ max ord p2q.
d
d|ppp´1q
d”1mod2
Let n be the largest odd divisor of p´1. Then Lppq “ ord p2q and Lpp2q “
n
ord p2q“lcmpLppq,ord p2qq,wherelcmpa ,...,a qstandsforthelowestcom-
np p 1 n
mon multiple of a ,...,a .
1 n
From the above tables, we guess there are formulas for Lppnq for ně2.
Proposition 4.6. Let p be a prime. Then we have
Lpp2qďpp´1qLppq.
8
Proof. Suppose p is a prime. Similar to the Proposition 4.1, we consider the
equation
x2m ”x pmod p2q,
and find the nontrivial solution for the maximal m. Suppose x is coprime with
p. Then it is equivalent to
x2m´1 ”1 pmod p2q. (.)
By Lemma 3.4, there are gcdp2m´1,ϕpp2qq solutions. By Lemma 3.1, ϕpp2q“
pp´1qp. By Euler theorem, for every odd prime p, we have
2p´1 ”1 pmod pq.
Thus, p|p2p´1´1q for pą2. Hence
gcdp2m´1,pp´1qpqą1
ô gcdp2p´1´1,2m´1qą1 or gcdpp´1,2m´1qą1.
Henceforfindthe largestm,weonlyneedtoconsidergcdp2p´1´1,2m´1qą1
with měp´1. Which infers that m“hpp´1q, hě1.
On the other hand, we canprovedthat hďLppq. In fact Figure 1 describes
the relationship between modulo p and modulo p2. Suppose
x2 ”x pmod pq, x2 ”x1 pmod p2q.
1 2 1 2
Then there exists some integers s and t such that
x2 “x `sp, x2 “x1 `tp2.
1 2 1 2
Thus, we have
x ´x1 “ppps´tq.
2 2
It shows that x1 and x lie in the same vertical line in Figure 1. And x12 ”
2 2 2
x2 ”x pmod pq. Let x12 ”x1 pmod p2q. Then it infers that x1 ”x pmod pq.
2 3 2 3 3 3
Hence,eachx1 liesinthesamelineofx . Therefore,theelementsinthe biggest
i i
cycle all lie in the area bounded by dashed lines below. Because there are no
more than p´1 vertical lines, the total numbers of the elements in the biggest
cycle is less or equal to pp´1qLppq. It means that Lpp2qďpp´1qLppq.
Proposition 4.7. For ně3, we have
Lppnq“pn´2¨Lpp2q.
Proof. Byusingthesameidea(asillustratedinFigure1)intheproofofPropo-
sition 4.6, we know that if x2 ” x pmod p2q and x2 ” x1 pmod p3q, then
i j i j
x1 ” x pmod p2q. Thus, following the arguments, we get Lpp3q ď p¨Lpp2q.
j j
Note that the multiplication factor is p, not p´1.
9
`1 `2 x1 xLppq x2 xi `p´1 p
0
`1 `2 x12 `p´1 2p
p
ďpp´1qLppqelements
`1 `2 x11 x1i `p´1 kp
pk´1qp
`1 `2 x1Lppq `p´1 pp´1qp
pp´2qp
`1 `2 `p´1 p2
pp´1qp
Figure 1: mod p and mod p2
If the largest cycle modulo p3 is less than pLpp2q. That is, the largest cycle
doesn’t take all of the elements in the area similar in Figure 1. Then, there
must be another largest cycle modulo p3. If use curve to describe the cycle,
thenthecurveintersecteachverticalline once. Theyallprojectontothe ”line”
r0,p2s. And they are not intersect (i.e., have no common elements). Since
Lpp3q ą Lpp2q, the largest cycle’s length must greater than Lpp2q. And note
that there are p rows(p is a prime number), then the number of largest cycles
must be 1. That is Lpp3q“pLpp2q.
For the general cases, the idea is the same. By conduction we complete the
proof.
Finally, we have
Proposition 4.8. Suppose m is a composite number, and m “ ps1ps2¨¨¨pst.
i1 i2 it
Here p are all prime numbers. Then we have
ik
Lpmq“Lpps1ps2¨¨¨pstq“lcm Lpps1q,Lpps2q,...,Lppstq .
i1 i2 it i1 i2 it
´ ¯
Proof. Let m“psipsj. Consider equation
i j
x2m ”x pmod psipsjq,
i j
where px,p q“px,p q“1. Then it is equivalent to the equation
i j
x2m´1 ”1 pmod psipsjq.
i j
By Lemma 3.4, it has gcdp2m´1,ϕppsipsjqq solutions.
i j
ϕppsipsjq“ϕppsiqϕppsjq“pp ´1qpsi´1¨pp ´1qpsj´1.
i j i j i i j j
Hence,wehaveLppsipsjq“lcmpLppsiq,Lppsjqq. Byconductionwewillcomplete
i j i j
the proof.
10
