Square-Congruence Modulo n

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1 Square-Cogruece Modulo Abstract This paper is a ivestigatio of a equivalece relatio o the itegers that was itroduced as a exercise i our Discrete Math class. Part I - Itro Defiitio Two itegers are Square-Cogruet Modulo if they are related to each other by the differece of squares relatio S o defied as as b if a b Sice S is a equivalece relatio o, the equivalece classes of S iduce a partitio of Defie to be that partitio. ={equivalece classes iduced by S } The Discrete Math exercises that started this ivestigatio were to fid the partitio for various values of. As a istructor, I wated to keep the problems maageable. I did t wat the umber of equivalece classes to be too great. So, I eeded a way to "predict" the umber of equivalece classes for ay. I eeded to kow the followig fuctio. Defiitio The fuctio u is : ( ) the umber of equivalece classes iduced by the size of the set Our motivatig task became to determie the values of first few values of. We did't otice a patter util 5 ad 6. S. Table 1 shows a summary for the Figure shows how for 5 the umbers that relate to 1 are all 1 less or 1 greater tha every multiple of 5. Ad the umbers that relate to are all less or greater tha every multiple of 5. Figure 3 shows a similar arragemet for 6. From Figures ad 3, two thigs are immediately apparet. The diagrams are periodic with period ad they are evely symmetric about. Because of these properties, to cout the equivalece classes you eed oly cosider the first few bracketed umbers that establish the First Half Cycle. 1

2 Accordig the patter established by Figures ad 3, the value of Half Cycle. This suggests that the sequece of values for But 4 does't fit the patter. is the size of 's First would be 1, odd 1,,,3,3,4,4,... 1, eve 4 is equal to istead of 3. We wodered if 4 was special because it's square or perhaps because it's a power of. But it turs out 4 is special for a more subtle reaso. Table ,0,1,,3,4,5, ,0,,4,6, ,1,3,5,7, ,0,3,6,9, ,1,,4,5,7,8, ,0,,4,6, ,1,3,5,7, ,0,5,10,15, ,1,4,6,9,......,,3,7,8, ,0,6,1,18, ,1,5,7,11,......,,4,8,10, ,3,9,15,1,... Figure Figure 3 It was clear that we eeded more empirical evidece to guide our cojectures. We made a catalog of the diagrams from 1 to 40. Table 4 shows the First Half Cycles for each. The First Half Cycles correspod to the bracketed umbers from Figures ad 3. The First Full Cycles are show i Table 5 for referece. Oe thig to otice right away is that if d is divisible by 4, its period is cut i half. This is the subtle reaso why 4 eds up beig special.

3 If has a First Half Cycle where every bar is touchig the diagoal, the cotais its maximum umber of equivalet classes. I that case we call complete. But ot all the Half Cycles are like that. Some Half Cycles have dropouts. A dropout is where the bar does ot reach the diagoal. Every dropout elimiates oe possible equivalece class ad reduces the size of by oe. I other words, the size of 's half cycle the umber of dropouts for the umber of bars that touch the diagoal Table 4 - Half Cycles Table 5 - Full Cycles 3

4 Now the challege becomes to fid patters amog the dropouts. If the dropouts have reliable patters, we may get a reliable way to compute. For example, it appears that all the prime umbers are complete ad have o dropouts. However, the primes are ot the oly complete umbers. Is there a overall patter for the complete umbers? Take a momet ad look for other patters. Please visit our website at to see the diagrams up closer. Part II - A Diagram Fuctio ad a Cap o If you take ay diagram ad fully exted it left ad right, you'll see what appears to be the graph of a eve, periodic fuctio o. This is the fuctio that takes each iteger to its equivalece class uder S where each equivalece class is represeted by its smallest oegative member o the vertical axis. We ca write this fuctio as Defiitio For the relatio S, the fuctio is :, ( a ) the smallest o egative iteger related to a Let's set up a few covetios before we go o. We defie the atural umbers to iclude 0,1,,3,... 1,,3,... for the positive zero,. Whe we wat to exclude zero, we ll use itegers. All throughout this paper ad ab, eve if ot explicitly stated. Theorem 1 - The fuctio (i) [ a] [ a] (ii) a is the smallest o egative member of (iii) [ a ] represets [ ] (iv) a b if ad oly if [ a] [ b] (v) (vi) sizeimg (i) Follows by defiitio sice a relates to a. a uiquely by its smallest o-egative member. a. (ii) Follows sice a is the set of all itegers related to a. (iii) Follows because the smallest o-egative member of a set of itegers is uique. 4

5 (iv) Assume [ a] [ b]. The a ad the same set. Hece, they must be the same. Coversely, if a b the a b, ad so by (i) [ a ] [ b ]. (v) We must show a a. Sice a a, by (ii) egative member of a. But by (i) [ ] [ ] a a. So a is the smallest o egative member of a which by (ii) is a. Hece a a. (vi) Defie the fuctio zeta : img, c c. We claim is bijective. Let cd, img such that c d. Sice c ad d are i the image of, by (v) c c ad d d. So, by (iv), we have c d ad is oe-to-oe. Now let c. The by (i), c c ad is oto. Therefore, is bijective ad we have size img size b are both the smallest o-egative member of a is the smallest o The ext theorem establishes the symmetric ad periodic properties of the diagrams that we observed previously. To prove the theorem, we'll rely o this lemma from Theorem 1 (iii). lemma a b Theorem - Properties of (i) (ii) (i) Sice if ad oly if asb. is eve: a a. is periodic with period : a a. has period a a (iii) If is divisible by 4, the a a 0 ad 0 established. (ii) Sice, ( a ) a a a a a we have that a a ( a ).. Hece as a ad (i) is we have that ( a ) a. Hece ( a ) S a ad (ii) is established. (iii) Let be such that 4, k k. We must show ( a ) S a. Sice 5

6 ( a ) a ( a k) a a 4k 4k a k k k k k we have that ( a ) a. Hece ( a ) S a ad (iii) is established. Sice divisible by 4 is special, we make these defiitios. Defiitio A eve umber divisible by 4 is called double-eve. A eve umber ot divisible by 4 is called sigle-eve. Furthermore, let E = {all sigle-eve umbers} S We offer these observatios without proof: (i) If is double eve, the is the product of two eves. (ii) All itegers belog i oe of three classes: odd {k1, k}, sigle-eve E {4k, k}, or double-eve {4 k, k}. s (iii) Sice all odd umbers are already the product of two odds, we do t defie double-odd. (iv) A sigle-eve umber caot be expressed as the product of two eves or two odds. Theorem ca ow be summarized as (i) is eve (ii) (iii) has period has period whe is double-eve Now we re ready to state the theorem that provides a upper boud for boud is simply the size of the First Half Cycle of.. This upper Theorem 3 - Cap o ( 1) /, odd ( ) +1, sigle eve 4 +1, double eve Sice is periodic o with period, it is oto the image of ay set of cosecutive itegers. For eve, choose the set of cosecutive itegers 1,...,0,...,. For 6

7 odd, choose the set 1,...,0,..., 1 of cosecutive itegers. Sice is a eve fuctio, we ca fold these sets across zero ad coclude that is oto the image of 0,..., if is eve ad is oto the image of 0,..., 1 whe is odd. If is double-eve, the has period. By the same argumet used for eve, we coclude that is oto the image of 0,..., 4 whe is double-eve. From Theorem 1 we kow that size img. Sice is a fuctio from to, the image of a fiite set uder must be the same size or smaller tha the set itself. So, for eve size img size 0,..., size 0,..., 1, for odd sizeimg size 0,..., 1 size 0,..., ad for double-eve size img size 0,..., 4 size 0,..., Part III - Completeess ad Dropouts The cap o i Theorem 8 tells us a lot about value is oe less tha the cap.. Let s defie a ew fuctio whose Defiitio ( 1) /, odd :, ( ), sigle eve 4, double eve Theorem 3 says that which meas that ( ) ( ) 1 max size of ( ) 1 Furthermore, the proof of Theorem 3 shows how the equivalece classes i amog the values of a 0,1,,...,. a o its First Half Cycle are foud Defiitio 7

8 We say that is complete if ( ) ( ) 1. If is complete, the all the bars i its First Half Cycle touch the diagoal. If is ot complete, the has dropouts where bars do ot touch the diagoal. We defie such bars as dropouts whe they occur i the First Half Cycle. Defiitio has a [a]-dropout if 0 a a. We say that has a dropout at a. a drops out to a ad that If has a [a]-dropout the the bar at a "drops out" to the height a. Ofte we oly eed to kow if has a dropout at a without eedig to kow what value it drops out to. For this, we have The Dropout Test. Theorem 4 - The Dropout Test has a [a]-dropout if ad oly if b a ad 0 b a for some umber b. The forward implicatio is clear from the defiitio. Cosider the coverse. Sice b is related to a, by the defiitio of we have 0 a b. Therefore 0 a a ad we have a [a]-dropout. We ow list some of the cosequeces of completeess. Theorem 5 - Completeess Whe is complete, (i) has the maximum umber of equivalece classes allowed by Theorem 3. (ii) { [0], [1], [],..., [ ( )] }. (iii) has o dropouts. Needs a little proof use Theorems 1 ad 3. By defiitio, if is complete, we kow how to compute. So, what umbers are complete? The previous cojecture that primes are complete is fairly easy to prove. Theorem 6 - p is complete If p is prime, the p is complete. 8

9 Evidetly, is complete. Let p be prime. Assume there is a dropout at a. We must show a cotradictio. By The Dropout Test, for some b we have p 1 b a 0 b a p. Therefore 0 a b p p a b This cotradicts b ad 0 a b p ad ad p a b p a ba b a b a is ot related to b. a. We'd like to "verify" Theorem 6 by checkig it agaist the diagrams. The Dropout Tree i Table th 6 summarizes the diagrams. The row lists the values of 0 a. The row a for values are the heights of the bars i the First Half Cycle for. The dropouts show up as colors differet from earby umbers. The complete umbers appear i The Dropout Tree as rows with the cosecutive umbers 0,1,,..., ad o discotiuous colors. Verify that all the primes look complete. Verify that there are complete umbers that are ot prime. What is the patter? Ca we characterize the umbers that are complete? Take a momet to otice other patters i The Dropout Tree. Visit our website at to see The Dropout Tree up to 100. Part IV - Characterizig [0] Let's take a iterlude to look at aother area of iterest. Ca we characterize the equivalece classes themselves? We begi by lookig at the zero class 0 which cotais all itegers equivalet to zero uder the square-cogruece relatio Theorem 7 - Characterizatio of [0] The zero class [0] is the set of all multiples of the smallest positive iteger whose square is divisible by. We eed some ew defiitios ad a ew fuctio to help us talk about Theorem 7. Defiitio If is composite but has o multiple prime factors, the is simple-composite. Otherwise, if is composite with multiple prime factors, the is multi-composite. S. 9

10 All itegers are oe of three thigs: prime, simple-composite, or multi-composite. All doubleeve umbers are multi-composite. Table 6 - The Dropout Tree up to =40 Defiitio The square-root prime-factor reductio fuctio is spr:, Theorem 8 - spr fuctio (i) spr ( ) m1 m m k 1 m1 m spr( ) p p p k whe p1 p p k (ii) spr ( ) is the smallest square divisible by. (iii) spr( ) is the smallest positive iteger whose square is divisible by. (iv) a if ad oly if spr( ) a. (v) spr( ) if ad oly if is prime or simple-composite. (vi) spr( ) if ad oly if is multi-composite m k 10

11 Needs a proof. Will hige o the fact that mi m i if ad oly if mi 1. Theorem 7 ca ow be restated ad proved. Theorem 7 - Characterizatio of [0] The zero class [0] for S is the set of all multiples of spr. Needs a proof. Should be straightforward usig the properties of spr. Corollaries to Theorem 7 (i) [0] cotais oly the multiples of iff is prime or simple-composite. (ii) [0] cotais more tha the multiples of iff is multi-composite. Theorem 7 leads to our first major result cocerig patters of dropouts. By the Dropout Test of Theorem 4, we have that 0 d 0 d. has a [0]-dropout iff for some d such that Theorem 9 - The [0]-dropout Theorem has [0]-dropouts if ad oly if is oe of three thigs: (i) odd multi-composite (ii) sigle-eve multi-composite (iii) double-eve where 4 is multi-composite. Needs a proof. Use Theorems 7 ad 8. So, oly multi-composite umbers have [0]-dropouts, but ot every multi-composite umber does. You ca check that this is true o the Dropout Tree i Table 6. The first few multicomposite umbers are multi-composites 8,9,1,16,18,0,4,5,7,8,3,36,40,... Multi-composites that do ot have [0]-dropouts are accouted for by the exceptio i Theorem 9 (iii). For example, 4 3 has o [0]-dropouts sice which is ot multicomposite. Although The [0]-dropout Theorem tells us whe has [0]-dropouts, it does't tell us how may [0]-dropouts there are. Coutig the umber of dropouts is crucial to computig. This topic will be explored later. 11