Everything You Need to Know About Modular Arithmetic

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Everything You Need to Know About Modular Arithmetic

A number a has an inverse modulo 26 if there is a b such that a·b ≡ 1(mod 26)or a·b = 26·k +1 thus we are looking for numbers whose products are 1 more than a multiple of 26 We create the following table Table 2: inverses modulo 26 x 1 3 5 7 9 11 15 17 19 21 23 25 x−1 (MOD m) 1 9 21 15 3 19 7 23 11 5 17 25

Solution: x y p x - UC Santa Barbara

≡ 1 (mod 23), and so 3 is a quadratic residue modulo 23 For p = 31, we have 3(31−1)/2 = 315 ≡ −1 (mod 31), and so 3 is a quadratic non-residue modulo 31 (3) Given that a is a quadratic residue modulo the odd prime p, prove the following: (a) a is not a primitive root of p (b) The integer p − a is a quadratic residue or non-residue

MATH 115A SOLUTION SET V FEBRUARY 17, 2005 Solution

(c) Using the exponents 2, 11 and 22, and working modulo 23 gives 22 ≡ 4, 211 ≡ 1 32 ≡ 9, 311 ≡ 1 52 ≡ 2, 511 ≡ 22, 522 ≡ 1 Thus 2, 3, and 5 have orders 11, 11 and 22 respectively (2) Establish each of the following statements below: (a) If a has order hk modulo n, then ah has order k modulo n

Primitive Roots mod p - homepagesmathuicedu

5 is a primitive root mod 23 It can be proven that there exists a primitive root mod p for every prime p (However, the proof isn’t easy; we shall omit it here ) 3) For each primitive root b in the table, b 0, b 1, b 2, , b p − 2 are all distinct in Z p, and they constituted all the nonzero elements of Z p Again, this is always true

18781 Solutions to Problem Set 4, Part 2

311 33 33 33 9 43 9 5 9 1 (mod 23); so 3 doesn’t work either 511 (52)5 5 25 5 9 5 1 (mod 23) and 52 2 (mod 23), so 5 is a primitve root mod 23 Now by the proof of existence of primitive roots mod p2, using Hensel’s lemma, only one lift of 5 will fail to be a primitive root mod 232:We need to check whether 522 1 (mod 232): 522 = (55)4 52

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Homework 9 Name - California State University, Fresno

116 - Number Theory Spring 2007 Homework 9 Name: Instructor: Travis Kelm Section 9 2 - Primitive Roots for Primes 1 Order Two: (a) Show that for any positive integer m > 2 we have that ordm(m −1)=2

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Everything You Need to Know About Modular Arithmetic...

Math 135, February 7, 2006

DefinitionLetm >0 be a positive integer called themodulus. We say that two integersaandbare congruent modulomifb-ais divisible bym. In other words, a≡b(modm)??a-b=m·kfor some integerk.(1) Note:

1. The notation ??≡??(modm) works somewhat in the same way as the familiar ?? =??.

2.acan be congruent to many numbers modulomas the following example illustrates.

Ex. 1The equation

x≡16(mod10) has solutionsx=...,-24-14,-4,6,16,26,36,46.... This follows from equation (1) since any of these numbers minus 16 is divisible by 10. So we can write x≡ ··· -24≡ -14≡ -4≡6≡16≡26≡36≡46(mod10). Since such equations have many solutions we introduce the notationa(MODm)

DefinitionThe symbol

a(MODm) (2) denotes the smallest positive numberxsuch that x≡a(modm). In other words,a(MODm) is the remainder whenais divided bymas many times as possible. Hence in example 1 we have

6 = 16(MOD10) and 6 =-24(MOD10) etc....

Relation between "x≡bmodm" and "x=bMODm"

x≡bmodmis an EQUIVALENCE relation with many solutions forxwhilex=bMODmis an EQUALITY. So one can think of the relationship between the two as follows x=b(MODm) is the smallest positive solution to the equationx≡b(modm). Since

0< b(MODm)< m

it is convention to take these numbers as the representatives for the class of numbersx≡b(modm).

1 Ex. 2The standard representatives for all possible numbers modulo 10 are given by

0,1,2,3,4,5,6,7,8,9

although, for example, 3≡13≡23(mod 10), we would take the smallest positive such number which is 3.

Inverses in Modular arithmetic

We have the following rules for modular arithmetic: Sum rule:IFa≡b(modm) THENa+c≡b+c(modm).(3) Multiplication Rule:IFa≡b(modm) and ifc≡d(modm) THENac≡bd(modm).(4) DefinitionAn inverse toamodulomis a integerbsuch that ab≡1(modm).(5) By definition (1) this means thatab-1 =k·mfor some integerk. As before, there are may be many

solutions to this equation but we choose as a representative the smallest positive solution and say that the

inversea-1is given by a -1=b(MODm). Ex 3. 3 has inverse 7 modulo 10 since 3·7 = 21 shows that

3·7≡1(mod 10) since 3·7-1 = 21-1 = 2·10.

5 does not have an inverse modulo 10. If 5·b≡1(mod 10) then this means that 5·b-1 = 10·kfor some

k. In other words

5·b= 10·k-1 which is impossible.

Conditions for an inverse ofato exist modulom

DefinitionTwo numbers are relatively primeif their prime factorizations have no factors in common. number modulom). Thenahas a multiplicative inverse modulomifaandmare relatively prime.

Ex 4Continuing with example 3 we can write 10 = 5·2. Thus, 3 is relatively prime to 10 and has an inverse

modulo 10 while 5 is not relatively prime to 10 and therefore has no inverse modulo 10.

Ex 5We can compute which numbers will have inverses modulo 10 by computing which are relatively prime

to 10 = 5·2. These numbers arex= 1,3,7,9. It is easy to see that the following table gives inverses module

10: 2

Table 1: inverses modulo 10

x1379 x -1MOD 101739

Ex 6: We can solve the equation 3·x+ 6≡8(mod 10) by using the sum (3) and multiplication (4) rules

along with the above table:

3·x+ 6≡8(mod 10) =?

3·x≡8-6≡2(mod 10) =?

(3 -1)·3·x≡(3-1)·2(mod 10) =? x≡7·2(mod 10)≡14(mod 10)≡4(mod 10) Final exampleWe calculate the table of inverses modulo 26. First note that

26 = 13·2

so that the only numbers that will have inverses are those which are rel. prime to 26...i.e. they contain no

factors of 2 or 13:

1,3,5,7,9,11,15,17,19,21,23,25.

Now we write some multiples of 26

26,52,78,104,130,156,182,208,234...

A numberahas an inverse modulo 26 if there is absuch that a·b≡1(mod 26)ora·b= 26·k+ 1.

thus we are looking for numbers whose products are 1 more than a multiple of 26. We create the following

table

Table 2: inverses modulo 26x1357911151719212325

x -1(MODm)1921153197231151725 since (using the list of multiples of 26 above)

1·1 = 1 = 26·0 + 1

3·9 = 27 = 26 + 1

5·21 = 105 = 104 + 1

7·15 = 105 = 104 + 1

11·19 = 209 = 208 + 1

17·23 = 391 = 15·26 + 1

25·25 = 625 = 26·24 + 1.

So we can solve

y= 17·x+ 12(MOD 26) forxby first considering the congruence equation y≡17·x+ 12(mod 26) and performing the following calculation (similar to ex 6) using the above table: y≡17·x+ 12(mod 26) =? y-12≡17·x(mod 26) =? (17 -1)(y-12)≡(17-1)·17·x(mod 26) =? (23)(y-12)≡(23)·17·x(mod 26) =?

23·(y-12)≡x(mod 26)

We now writex= 23·(y-12)(MOD 26).

The difference between

23·(y-12)≡x(mod 26)

and x= 23·(y-12)(MOD 26)

is simply that in the first equation, a choice ofywill yield many different solutionsxwhile in the second

equation a choice ofygives the valuexsuch thatxis the smallest positive solution...i.e. the smallest positive

solution to the first equation. 4quotesdbs_dbs8.pdfusesText_14

[PDF] Everything You Need to Know About Modular Arithmetic

Math 135, February 7, 2006

1. The notation ??≡??(modm) works somewhat in the same way as the familiar ?? =??.

2.acan be congruent to many numbers modulomas the following example illustrates.

Ex. 1The equation

DefinitionThe symbol

6 = 16(MOD10) and 6 =-24(MOD10) etc....

Relation between "x≡bmodm" and "x=bMODm"

0< b(MODm)< m

0,1,2,3,4,5,6,7,8,9

Inverses in Modular arithmetic

3·7≡1(mod 10) since 3·7-1 = 21-1 = 2·10.

5 does not have an inverse modulo 10. If 5·b≡1(mod 10) then this means that 5·b-1 = 10·kfor some

5·b= 10·k-1 which is impossible.

Conditions for an inverse ofato exist modulom

Table 1: inverses modulo 10

3·x+ 6≡8(mod 10) =?

3·x≡8-6≡2(mod 10) =?

26 = 13·2

1,3,5,7,9,11,15,17,19,21,23,25.

Now we write some multiples of 26

26,52,78,104,130,156,182,208,234...

Table 2: inverses modulo 26x1357911151719212325

1·1 = 1 = 26·0 + 1

3·9 = 27 = 26 + 1

5·21 = 105 = 104 + 1

7·15 = 105 = 104 + 1

11·19 = 209 = 208 + 1

17·23 = 391 = 15·26 + 1

25·25 = 625 = 26·24 + 1.

So we can solve

23·(y-12)≡x(mod 26)

We now writex= 23·(y-12)(MOD 26).

The difference between

23·(y-12)≡x(mod 26)