Indices and Applications 3

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Indices and Applications

Justin Stevens

January 28, 2013

Abstract

Indices are one of the most overlooked but useful parts of NumberTheory. Indices are rarely known, so in this article I would like to takethe time to introduce them. The background knowledge you need forthis is primitive roots and order. Without these concepts, you willprobably get confused throughout this article. Many problems andideas throughout this article are inspired by [1] and problems and orsolutions from [1] are sourced accordingly. The core of this articlerelies on the problems the reader attempts to do to grasp the concept.Many of the theorems presented in this article were confusing to meuntil I tried problems involving the theorems and it suddenly madesense. Therefore, the theory in this article is limited to a couple ofpages to provide you the necessary knowledge while there is quite afew problems at the end.

Many thanks to Evan Chen for help with LATEX. Also, many thanks to Kyle Gettig

for finding many errors in this document.

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Contents

1 Order and Primitive roots 3

2 What are indices? 5

3 Some extensions 6

4 Exercises for the reader 8

5 Solutions 9

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1 Order and Primitive roots

Before I get into a lot of stuff about indices, I want to use primitive rootsand order in a few problems as indices strongly rely on order and primitiveroots.

Example 1 ([4]). Let p be a prime. Let g be a primitive root of modulo p.Prove there is no k such that gk+2 gk+1 + 1 gk + 2 (mod p)

Proof. If gk+2 gk+1 + 1 (mod p), then gk+1(g 1) 1 (mod p) and thesecond gives gk(g 1) 1 (mod p). Note that since gcd(g 1, p) = 1 (fromthese congruences), and gcd(gk, p) = 1 (from congruences), substitute a =gk, b = gk+1, and x = g 1. Therefore, ax bx (mod p) = (a b)x 0

(mod p), but gcd(x, p) = 1 = a b 0 (mod p), or gk gk+1 (mod p).However, this implies gk(g 1) 0 (mod p), absurd.

This problem didnt really utilize primitive roots; the next one will.

Example 2 ([4]). Prove that the product of all primitive roots ofp between1 and p 1 where p > 3 is a prime is 1 mod p.

Proof.

Lemma 1 ([1]). Prove that if r is a primitive root of the odd prime p, and

rr

1 (mod p), then r

is a primitive root of the odd prime p.

Proof. Note that (rr)d 1 (mod p) for all d. Assume rd 1 (mod p) forsome d < (p) = p 1. However, then rd 1 (mod p), contradicting rbeing a primitive root mod p. Therefore, r is a primitive root ofp also.

Therefore, we pair all primitive roots into (p1)2 groups (which is aninteger) to get a product of 1. The only other thing we have to note is that1 and p 1 both cant be a primitive root of p because p > 3.

Example 3 ([5]). Let m = (4p 1)/3, where p is a prime number exceeding3. Prove that 2m1 has remainder 1 when divided by m.

Proof. This condition is the same as om(2)|m1. Note that 4p 1 (mod m)

therefore om(2)|2p. We now want 2p|m 1 or 2p|4p43 or 2p|4

p 4. Notethat 4p 4 0 (mod 2) and 4p 4 0 (mod p) from Fermats, thereforeby Chinese Remainder Theorem and gcd(2, p) = 1 we are done.

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Example 4 ([4]). Let g be a Fibonnaci primitive root i.e a primitive root

that satisfies g2 g + 1 (mod p) where p is an odd prime. a. Prove thatg 1 is also a primitive root of p. b. Ifp is of the form 4k + 3, prove that(g 1)2k+3 g 2 (mod p), and decude that g 2 is a primitive root of p.

Proof. a. We get g(g 1) 1 (mod p) from Fibonnaci primitive root,therefore, using the above lemma, g 1 is also a primitive root of p.

b. From the first part g 1 is a primitive root of p, therefore (g 1)2k+1 1 (mod p). Therefore, we want (g 1)2 g 2 (mod p) or(g2 2g + 1) g 2 (mod p). But g2 g + 1 so this is true.

Now, let d = op(g 2). Therefore, (g 2)d (g 1)d(2k+3) 1

(mod p) (p)|d(2k + 3). This happens when p 1|d(2k +3). However,

gcd(p 1, 2k + 3) = 1 because p = 4k + 3, therefore p 1|d and g 2 is aprimitive root of p.

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2 What are indices?

Let r be a primitive root (mod n). A well-known theorem states that theset {r1, r2, , r(n)} and the set of integers relatively prime to n are in aone to one correspondence and contain the same elements. Let gcd(b, n) = 1.Then, the smallest positive integer a such that ra b (mod n) is defined asindr(b).

As an example, 3 is a primitive root mod 7 (easy to check). Lets saywe want to compute the smallest integer such that 3a 5 (mod 7) (or inother terms, ind3(5)). To do this is a matter of basic computation, 3

1 3(mod 7), 32 2 (mod 7), 33 6 (mod 7), 34 4 (mod 7), 35 5 (mod 7),so the answer is 5.

A small question the reader may have is why r must be a primitive root(mod n) to imply the existence of an index? Assume that r has order gwith g < (n). The set {r1, r2, r3, , rg} make up all possible values of b,which has cardinality g. But the number of possible values for b is (n) andg < (n), so we have arrived at a contradiction.

As an example of this, 2 has order 3 mod 7. Therefore, note that21 2 (mod 7), 22 4 (mod 7), 23 1 (mod 7), 24 2 (mod 7), 25 4(mod 7), 26 1 (mod 7). Therefore, we cannot find ind2(x) for x = 3, 5, 6.

Note that 2 is a primitive root of 9. ind2(2 5) = ind2(10) which is thesmallest value such that 2x 1 (mod 9) or x = 0. Also, ind2(2) = 1 andind2(5) = 5, and ind2(2) + ind2(5) ind2(10) (mod (9)).

The following theorems are very fundamental theorems in indices

Theorem 1 ([1]). Letr be a primitive root ofn. Then, indr(ab) indr(a)+indr(b) (mod (n)). Also, indr(a

k) kindr(a) (mod (n)).

Proof Outline. Both follow from the fact that if r is a primitive root of nand rx ry (mod n), then x y (mod (n)). Try filling in the detailsyourself, however, if you get stuck, take a look at the solution to Problem3. [1]

The true importance of these theorems will become apparent in the nextsection.

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3 Some extensions

Example 5. Find all solutions of x3 4 (mod 9)

Solution. We first find a primitive root of 9. After quick computations, 2 isindeed a primitive root of 9. Therefore, we can take the index base 2 of bothsides to give us 3ind2(x) ind2(4) (mod (9)). Obviously, ind2(4) = 2,therefore, 3ind2(x) 2 (mod 6), which implies that this equation has nosolutions.

Example 6. Find all solutions of x3 8 (mod 9).

Solution. x = 2 is an obvious solution, let us find all solutions using indices.

Again, 2 is a primitive root of 9, so therefore taking ind2, we get 3ind2(x) ind2(8) (mod (9)). Without doing much computation, ind2(8) = 3, there-fore we get 3ind2(x) 3 (mod 6) = ind2(x) 1 (mod 2). Sinceindr(x) (n) where r is a primitive root of n, we are going to getind2(x) = 1, 3, 5. The first case gives x 2

1 (mod 9) and the second givesx 23 8 (mod 9) and the third gives x 25 32 5 (mod 9) all ofwhich work.

From these two examples, we have a path to prove the following twotheorems.

Theorem 2. The equation xk a (mod n) where there exists a primitive

root of n has a solution iff gcd(k, (n))|indr(a)| where r is a primitive rootof n. [1]

Proof Sketch. The proof of this follows from our two examples above usingsimilar methods.Try filling in the details yourself, however, if you get stuck,take a look at the solution to Problem 3. [1]

Theorem 3. The equation xk a (mod n) where there exists a primitive

root of n has a solution iff a(n)

gcd((n),k)) 1 (mod n). [1]

Proof Sketch. The proof of this follows from Theorem 2 and taking indr

where r is a primitive root of n. Try filling in the details yourself, however,if you get stuck, take a look at the solution to Problem 3. [1]

Example 7. Does there exist an integer x such that x8 9 (mod 11)?

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Solution. 9(11)

gcd((11),8) 95 (81)2 9 45 (mod 11) 1 (mod 11), sosolutions do exist. The nice thing about this theorem is that in Theorem2 we have to compute a primitive root of 11 (thankfully 2 works) but forlarger numbers it is a tedious work. Using this method, we get a methodthat involves just using the numbers given to us and computation.

Theorem 4. Prove a more general version of the first part of theorem 1,that is, prove that if n has a primitive root r, then indr(a1 a2 ak) indr(a1) + indr(a2) + indr(ak) (mod (n)) where k 2 is an integer.

Proof. Assume it holds for k = p and I prove it holds for k = p + 1 with basecase k = 2 being Theorem 1. indr((a1 a2 ak)(ak+1)) indr(a1 a2

ak)+indr (mod (n)) from Theorem 1 then use the inductive hyptothesisand the rest is obvious.

Example 8. Compute 3501 5221 7318 (mod 17)

Solution. First off, 3 is a primitive root of 17 (tedious to calculate but(17) = 16 makes it only a matter of calculating 38 (mod 17)). Therefore,let x = 3501 5221 7318, and we get ind3(x) 501+221ind3(5)+318ind3(7)(mod 16).

Using an index table, we find ind3(5) = 5 and ind3(7) = 11 (note thatthese could be done by hand too, however, with ugly numbers like this indextables are nice). Therefore, we get ind3(x) 501+2215+31811 (mod 16)

or therefore is 5 + 1 + 10 = 16, therefore we get ind3(x) = 0 = x 1(mod 17).

While this approach is very computational, it is a method that helpswhen you have incredibly large numbers as bases (such as 101195 103159

(mod 163)). In this case, an index table is also nice to have.

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4 Exercises for the reader

As said before in the introduction, you will learn most from these problems.Try all the problems carefully, making sure you understand how to useindices on the problems. If you get stuck on any problems, solutions arein the next section but please try to solve the problem before looking atsolutions.

Problem 1. Solve the equation x5 1 (mod 9).

Problem 2. Why do our methods not work for the equation x5 2(mod 24)?

Problem 3. Fill in the details for our proofs of Theorem 1, Theorem 2, andTheorem 3 (the main idea of the proof is given).

Problem 4. Solve the equation x4 5 (mod 9) or prove it has no solutions.

Problem 5. Compute the remainder when 52997100 is divided by 9. (Usingindices)

Problem 6 ([1]). If p is a prime and gcd(k, p 1) = 1, prove that theintegers 1k, 2k, 3k, , (p 1)k form a reduced set of residues modulo p.

Problem 7 ([1]). If r and r are both primitive roots of the odd prime p,

show that for gcd(a, p) = 1, then indr(a) (indr(a))(indrr) (mod p 1).Problem 8 ([1]). Show that the congruence x2 + 1 0 (mod p) has asolution iff p is of the form 4k + 1.

Problem 9 ([1]). If r is a primitive root of the odd prime p, show thatindr(p a) indr(a) +

p12 (mod p).

Problem 10 ([1]). Let r be a primitive root of the odd prime, and letd = gcd(k, p 1). Prove that the values of a for which the congruence

xk a (mod p) is solvable are rd, r2d, , rp1d

d.

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5 Solutions

Problem 1. 2 is a primitive root of 9, therefore, take ind2. We end up with5ind2(x) 0 (mod 6) = ind2(x) = 1 or x 2 (mod 9).

Problem 2. There does not exist a primitive root of 24 because 24 = 23 3and is not one of the desired forms for primitive roots.

Problem 3. All of these proofs are from [1]Theorem 1: Note that rindr(ab) ab (mod n) and rindr(a)+indr(b)

rindr(a)rindr(b) ab (mod n) therefore because r is a primitive root of n,indr(ab) indr(a) + indr(b) (mod (n)) (well known theorem on primitiveroots).

Theorem 2: The equation xk

a (mod n) implies that taking indr, weget kindr(x) indr(a) (mod (n)) If gcd(k, (n))|indr(a) then we can finda sufficient x such that indr(x) satisfies this equation.

Theorem 3: The equation a(n)

gcd((n),k) 1 (mod n) is true iff indr(a) (n)

gcd((n),k) 0 (mod (n)) by taking indr where r is a primitive root of n,

which is the same as gcd((n), k)|indr(a) which is the same condition asTheorem 2, so we are done.

Problem 4. 5(9)

gcd((9),4) 562 53 7 (mod 9) so no solutions exist using

Theorem 3.

Problem 5. 2 is a primitive root of 9, so take the index base 2 to giveus ind2(5299 7100) 299ind2(5) + 100ind2(7) (mod 6). Note that 2x 5(mod 9) is solved when x = 5, therefore ind2(5) = 5 and 2

x 7 (mod 9) issolved when x = 4, therefore ind2(7) = 4. Therefore, we get (2995+1004)(mod 6) 5 (mod 6), or ind2(5

299 7100) = 5 = 5299 7100 5 (mod 9).

Problem 6. We desire to prove that ak bk (mod p) is impossible wherea and b are in {1, 2, 3, 4, , p 1}. Since p is a prime, a primitive rootexists, let it be r. Therefore, we get kindr(a) kindr(b) (mod p 1).Since gcd(k, p 1) = 1, we get indr(a) indr(b) (mod p 1). However,looking at the set {r, r2, , rp1} we note that this is the same as the set{1, 2, 3, , p1}, therefore there is only one corresponding value for indr(a)

and indr(b) and indr(a) = indr(b) a = b and we are done .Problem 7. We are going to use a similar method as in the outline totheorem 1. Note that (r)indra a (mod n). Also, note that (rindrr)indra rindra a (mod n). Therefore, (r)indra (r)indrrindra (mod n) and by awell-known theorem on primitive roots and the fact r is a primitive root ofn, this implies indra indrr indra (mod (n)).

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Problem 8. First off, for 4k+3, by Theorem 3, we must have (1)(p)

gcd((p),2) =1 (mod p) (we can use this theorem because there is a primitive root

(mod p) where p is prime). This happens only if (p)gcd((p),2) 1 (mod 2).

However, we have p1gcd(2,p1)

is equal to p12 (because p = 4k + 3) which is2k + 1 and 4k + 3 doesnt work.

Now, use theorem 3 for 4k + 1 to give us a desired (1)(p)

gcd((p),2)

1 (mod p). Note that this holds only if (p)gcd((p),2) and (p) = 4k and

gcd((p), 2) = 2, so therefore we get (p)gcd((p),2)

= 2k and we have provedit only works for 4k + 1 .

Problem 9. This condition is the same as rindr(pa) rindr(a)+p12 (mod p)

using the converse of theorem 1. The LHS is the same as p a and the RHSis the same as a (1) = a, so therefore we are done .

Problem 10. By Theorem 3, we must have a(p)d 1 (mod p). Therefore,

a = rd, r2d, , rp1d

d all work (should be obvious). Also, from Lagrange,

there are at most (p)d

solutions, and we have proven there are (p)d

= p1d

solutions above, therefore there are exactly (p)d

solutions which are the onesabove.

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References

[1] Burton, David M. Section 8.4. Elementary number theory. Boston:Allyn and Bacon, 1976.

[2] Primitive Roots and Indices. Primitive Roots and Indices. 22 Jan.2013http://aubuchon.jsc.vsc.edu/numberTheoryNotes16PrimitiveRootsIndices.htm.

[3] alozano. using primitive roots and index to solve con-gruences (version 1). PlanetMath.org. Freely availableathttp://planetmath.org/UsingPrimitiveRootsAndIndexToSolveCongruences.html.

[4] Peter Vandendriessche, and Hojoo Lee. Problems Elementary Number

Theory. http://www.math.muni.cz/ bulik/vyuka/pen-20070711.pdf.(Note: All these problems are from various sources which can be foundin this document, look at section 3.1)

[5] WOOT: Worldwide Online Olympiad Training. WOOT. 27 Jan. 2013http://www.artofproblemsolving.com/School/woot.php.

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