MATH 324 Elementary Number Theory Solutions to Assignment 5 Due: Wednesday August 10, 2005

Department of Mathematical and Statistical Sciences University of Alberta

Question 1. [p 345. #3]

Find the number of primitive roots of each of the following primes.

(a) 7, (b) 13, (c) 17, (d) 19, (e) 29, (f) 47.

Solution:

(a) p = 7, φ(

φ(7))

= φ(6) = φ(2) · φ(3) = 1 · 2 = 2.

(b) p = 13, φ(

φ(13))

= φ(12) = φ(3) · φ(4) = 2 · 2 = 4.

(c) p = 17, φ(

φ(17))

= φ(16) = φ(24) = 23 = 8.

(d) p = 19, φ(

φ(19))

= φ(18) = φ(2) · φ(32) = 1 · 3 · 2 = 6.

(e) p = 29, φ(

φ(29))

= φ(28) = φ(4) · φ(7) = 2 · 6 = 12.

(f) p = 47, φ(

φ(47))

= φ(46) = φ(2) · φ(23) = 1 · 22 = 22.

Question 2. [p 345. #8]

Let r be a primitive root of the prime p with p ≡ 1 (mod 4). Show that −r is also a primitive root.

Solution: Since p ≡ 1 (mod 4), then p is odd, and

rp−1 ≡ 1 (mod p) implies p∣

∣

(

rp−12 − 1

)(

rp−12 + 1

)

,

and ordp(r) = p − 1 implies that p∣

∣ rp−12 + 1, that is, r

p−12 ≡ −1 (mod p). Multiply this by r to get

rp+12 ≡ (−r)

p+12 ≡ −r (mod p).

Therefore,

ordp(−r) = ordp

(

rp+12

)

=p − 1

(

p − 1, p+12

) ,

and since p = 4k+1, then(

p−1, p+12

)

= (4k, 2k+1) = 1. Therefore, ordp(−r) = p−1, and −r is a primitiveroot modulo p.

Question 3. [p 345. #10]

(a) Find the number of incongruent roots modulo 6 of the polynomial x2 − x.

(b) Explain why the answer to part (a) does not contradict Lagrange’s theorem

Solution:

(a) There are 4 incongruent roots modulo 6 of the polynomial x2 − x, namely, x = 0, 1, 3, 4, since

for x = 0, x2 − x ≡ 0 (mod 6)for x = 1, x2 − x ≡ 0 (mod 6)for x = 2, x2 − x ≡ 2 (mod 6)for x = 3, x2 − x ≡ 0 (mod 6)for x = 4, x2 − x ≡ 0 (mod 6)for x = 5, x2 − x ≡ 2 (mod 6).

(b) This does not contradict Lagrange’s theorem, since the modulus 6 is not a prime, and Lagrange’stheorem does not apply.

Question 4. [p 354. #1]

Which of the integers 4, 10, 16, 22, 28 have a primitive root?

Solution: Recall that only the integers2, 4, pα, 2pα

where α ≥ 1 and p is an odd prime, have primitive roots, therefore, of the integers mentioned above, only4, 10 = 2 · 5, 22 = 2 · 11 have primitive roots.

Question 5. [p 354. #13]

Show that if the positive integer m has a primitive root then the only solutions of the congruence

x2 ≡ 1 (mod m)

are x ≡ ±1 (mod m).

Solution: Clearly, x ≡ 1 (mod m) and x ≡ −1 (mod m) are solutions to the quadratic congruencex2 ≡ 1 (mod m).

Now, if m has a primitive root, then m must be one of the following: 2, 4, pα, or 2pα where α ≥ 1 and p

is an odd prime.

case 1: m = 2.

Suppose that x = x0 is a solution to x2 ≡ 1 (mod 2), then 2∣

∣ (x0 − 1)(x0 + 1), and therefore, either

2∣

∣ x0 − 1 or 2∣

∣ x0 + 1, that is, either x0 ≡ 1 (mod 2) or x0 ≡ −1 (mod 2). So, the only solutions to

x2 ≡ 1 (mod 2) are x ≡ ±1 (mod 2), that is, x ≡ 1 (mod 2).

case 2: m = 4.

Suppose that x = x0 is a solution to x2 ≡ 1 (mod 4), then 4∣

∣ (x0 − 1)(x0 + 1), so that 2∣

∣ (x0 − 1)(x0 + 1),

and either 2∣

∣ x0 − 1 or 2∣

∣ x0 + 1, so that x0 must be odd. Therefore either x0 ≡ 1 (mod 4) or

x0 ≡ 3 ≡ −1 (mod 4). So, the only solutions to x2 ≡ 1 (mod 4) are x ≡ ±1 (mod 4).

case 3: m = pα, where α ≥ 1 and p is an odd prime.Suppose that x = x0 is a solution to x2 ≡ 1 (mod pα), then p

∣

∣ (x0 − 1)(x0 + 1) since pα∣

∣ (x0 − 1)(x0 + 1),

and since p is a prime, then either p∣

∣ x0 −1 or p∣

∣ x0 +1. Now, p cannot divide both x0 −1 and x0 +1, since

then it would divide 2 = (x0 + 1)− (x0 − 1), which can’t happen since p is odd. If p∣

∣ x0 − 1, and p∣

∣6 x0 + 1,

then pα∣

∣ x0 − 1 also. If p∣

∣ x0 + 1, and p∣

∣6 x0 − 1, then pα∣

∣ x0 + 1 also. Therefore, x ≡ ±1 (mod pα) are

the only solutions to the congruence x2 ≡ 1 (mod pα).

case 4: m = 2pα, where α ≥ 1 and p is an odd prime.Suppose that x = x0 is a solution to x2 ≡ 1 (mod 2pα), then 2

∣

∣ (x0−1)(x0 +1) since 2pα∣

∣ (x0−1)(x0 +1),

and x0 is odd. Also, the same argument as above shows that either pα∣

∣ x0 − 1 or pα∣

∣ x0 + 1. Therefore,

either 2pα∣

∣ x0 − 1 or 2pα∣

∣ x0 + 1, that is, the only solutions to x2 ≡ 1 (mod 2pα) are x ≡ ±1 (mod 2pα).

Note: According to the proof in the book: If r is a primitive root modulo m, then (r, m) = 1, so that theintegers

r1, r2, . . . , rφ(m)

form a reduced residue system modulo m. Now, if x2 ≡ 1 (mod m), then (x, m) = 1, so there exists a t

with 1 ≤ t ≤ φ(m), such that x ≡ rt (mod m). Therefore, r2t ≡ 1 (mod m), and since r is a primitive

root modulo m, then φ(m)∣

∣ 2t. Thus, there is an integer k such that t = kφ(m)2 , so that

x ≡ rt ≡ rkφ(m)

2 =

(

rφ(m)

2

)k

≡ (−1)k ≡ ±1 (mod m),

since rφ(m)

2 ≡ −1 (mod m). It is this last congruence that I am not sure about, he seems to be assuming

that m∣

∣ rφ(m) − 1 =(

rφ(m)

2 − 1)(

rφ(m)

2 + 1)

implies that m∣

∣ rφ(m)

2 + 1. But this is exactly what we aretrying to show. See the next problem . . .

Question 5a. [p 354. #14]

(Not Assigned) Let n be a positive integer possessing a primitive root. Using this primitive root, provethat the product of all positive integers less than n and relatively prime to n is congruent to −1 modulo n.

(When n is prime, this result is Wilson’s Theorem.)

Solution: If n = 2, the result is trivial, since 1 ≡ −1 (mod 2).

Suppose then that n > 2 and r is a primitive root modulo n, then the set

r, r2, . . . , rφ(n)

is a reduced residue system modulo n, and since rφ(n) ≡ 1 (mod n), then

n∏

k=1(k,n)=1

k ≡

φ(n)∏

k=1

rk ≡

φ(n)−1∏

k=1

rk ≡ r

φ(n)−1∑

k=1

k

≡ r12φ(n)[φ(n) − 1] (mod n),

so that

n∏

k=1(k,n)=1

k

2

≡ rφ(n)[φ(n)−1] ≡(

rφ(n))φ(n)−1

≡ 1 (mod n).

Now, ifn∏

k=1(k,n)=1

k ≡ 1 (mod n), then this implies that

φ(n)−1∏

k=1

rk ≡ r12φ(n)[φ(n) − 1] ≡ 1 (mod n),

so that ordnr = φ(n)∣

∣

12φ(n)[φ(n)−1], and 1

2φ(n)[φ(n)−1] = l ·φ(n) for some integer l, that is, φ(n)−1 = 2l.

But this is a contradiction, since φ(n) is even for n > 2. Therefore,

n∏

k=1(k,n)=1

k

2

≡ 1 (mod n)

butn∏

k=1(k,n)=1

k 6≡ 1 (mod n).

We know that all solutions of the congruence x2 ≡ 1 (mod n) are of the form rt for some integer t, andthe number of solutions to the congruence x2 ≡ 1 (mod n) is the same as the number of solutions tothe congruence r2t ≡ 1 ≡ r0 (mod n), which is the same as the number of solutions to the congruence2t ≡ 0 (mod φ(n)), which is d = (2, φ(n)) = 2. The solutions are x ≡ ±1 (mod n).

Therefore, the congruence x2 ≡ 1 (mod n) has only the solutions x ≡ ±1 (mod n), and since

n∏

k=1(k,n)=1

k 6≡ 1 (mod n),

then we must haven∏

k=1(k,n)=1

k ≡ −1 (mod n).

Note: In general, the congruence x2 ≡ 1 (mod n) has more than two solutions. In fact, if the prime powerdecomposition of n is given by

n = 2αpα11 pα2

2 · · · pαk

k

where p1, p2, . . . , pk are distinct odd primes and α ≥ 1, αi ≥ 1, then the congruence x2 ≡ 1 (mod n) hasexactly

(a) 2k incongruent solutions modulo n, if α = 0 or α = 1

(b) 2k+1 incongruent solutions modulo n, if α = 2

(c) 2k+2 incongruent solutions modulo n, if α > 2.

Question 6. [p 412. #5]

Evaluate the Legendre symbol

(

7

11

)

(a) Using Euler’s criterion.

(b) Using Gauss’s lemma.

Solution:

(a) Here a = 7, p = 11, so thatp − 1

2= 5, and Euler’s criterion says

(

a

p

)

≡ ap−12 (mod p)

so that

(

7

11

)

≡ 75 (mod 11).

Now,

71 ≡ 7 (mod 11), 72 ≡ 5 (mod 11), 73 ≡ 2 (mod 11),

74 ≡ 3 (mod 11), 75 ≡ −1 (mod 11),

so that

(

7

11

)

≡ −1 (mod 11), that is,

(

7

11

)

= −1.

(b) Again, a = 7, p = 11, p−12 = 5, and the set {1 · 7, 2 · 7, 3 · 7, 4 · 7, 5 · 7} = {7, 14, 21, 28, 35} has least

positive residues {7X, 3, 10X, 6X, 2} modulo 11 and exactly s = 3 of them are greater thanp

2=

11

2.

Therefore, by Gauss’ lemma,

(

7

11

)

= (−1)3 = −1.

Question 7. [p 412. #7]

Show that if p is an odd prime, then

(

−2

p

)

=

{

+1 if p ≡ 1 or 3 (mod 8)

−1 if p ≡ −1 or − 3 (mod 8)

Solution: Since(

−2

p

)

=

(

(−1) · 2

p

)

=

(

−1

p

) (

2

p

)

= (−1)p−12 · (−1)

p2−18 ,

then

(a) If p ≡ 1 (mod 8), say p = 8k + 1, then p−12 = 4k and p2

−18 = 8k2 + 2k, and therefore,

(

−2

p

)

= (+1)(+1) = +1.

(b) If p ≡ 3 (mod 8), say p = 8k + 3, then p−12 = 4k + 1 and p2

−18 = 8k2 + 6k + 1, and therefore,

(

−2

p

)

= (−1)(−1) = +1.

(c) If p ≡ −1 (mod 8), say p = 8k − 1, then p−12 = 4k − 1 and p2

−18 = 8k2 − 2k, and therefore,

(

−2

p

)

= (−1)(+1) = −1.

(d) If p ≡ −3 (mod 8), say p = 8k − 3, then p−12 = 4k − 2 and p2

−18 = 8k2 − 6k + 1, and therefore,

(

−2

p

)

= (+1)(−1) = −1.

Therefore,(

−2

p

)

=

{

+1 if p ≡ 1 or 3 (mod 8)

−1 if p ≡ −1 or − 3 (mod 8),

as stated.

Question 8. [p 412. #12a]

Consider the quadratic congruence ax2 + bx + c ≡ 0 (mod p), where p is prime and a, b, and c are integerswith p

∣

∣6 a.

Let p = 2. Determine which quadratic congruences (mod 2) have solutions.

Solution:

If 2∣

∣6 a, then the congruence ax2+bx+c ≡ 0 (mod 2) is equivalent to the congruence x2+bx+c ≡ 0 (mod 2).

If both b ≡ 0 (mod 2) and c ≡ 0 (mod 2), then x ≡ 0 (mod 2) is the only solution to x2 + bx+ c ≡ 0 (mod 2).

If b ≡ 0 (mod 2) and c 6≡ 0 (mod 2), then x ≡ 1 (mod 2) is the only solution to x2 + bx + c ≡ 0 (mod 2).

If b 6≡ 0 (mod 2) and c ≡ 0 (mod 2), then x ≡ 0 (mod 2) and x ≡ 1 (mod 2) are both solutions tox2 + bx + c ≡ 0 (mod 2).

If b 6≡ 0 (mod 2) and c 6≡ 0 (mod 2), then there are no solutions to x2 + bx + c ≡ 0 (mod 2).

Question 9. [p 413. #12b]

Consider the quadratic congruence ax2 + bx + c ≡ 0 (mod p), where p is prime and a, b, and c are integerswith p

∣

∣6 a.

Let p be an odd prime and let d = b2−4ac, Show that the congruence ax2 +bx+c ≡ 0 (mod p) is equivalentto the congruence y2 ≡ d (mod p), where y = 2ax + b. Conclude that if d ≡ 0 (mod p), then there isexactly one solution x modulo p, if d is a quadratic residue of p, then there are two incongruent solutions,and if d is a quadratic nonresidue of p, then there are no solutions.

Solution: Let p be an odd prime and let d = b2 − 4ac, since p∣

∣6 a, and p is odd, then (p, 4a) = 1, so x is a

solution to ax2 + bx + c ≡ 0 (mod p) if and only if x is a solution to the congruence 4a2x2 + 4abx + 4ac ≡0 (mod p). that is, if and only if x is a solution to the congruence (2ax + b)2 ≡ b2 − 4ac (mod p), that is, ifand only if y2 ≡ d (mod p), where y = 2ax + b and d = b2 − 4ac.

If b2 − 4ac is a quadratic residue modulo p, then there exists an r such that r2 ≡ b2 − 4ac (mod p) andthere are two incongruent solutions to ax2 + bx + c ≡ 0 (mod p).

If b2 − 4ac is a quadratic nonresidue modulo p, then no such r exists, so the congruence ax2 + bx + c ≡0 (mod p) has no solutions.

If b2 − 4ac ≡ 0 (mod p), then (2ax + b)2 ≡ 0 (mod p) implies 2ax + b ≡ 0 (mod p), and since (p, 2a) = 1,

this linear congruence has exactly one solution.

Question 10. [p 426. #13]

Find all solutions of the quadratic congruences

(a) x2 + x + 1 ≡ 0 (mod 7)

(b) x2 + 5x + 1 ≡ 0 (mod 7)

(c) x2 + 3x + 1 ≡ 0 (mod 7)

Solution:

(a) Here a = 1, b = 1, c = 1, and b2 − 4ac = −3 ≡ 4 ≡ 22 (mod 7), so that b2 − 4ac is a quadraticresidue modulo 7 and (2x + 1)2 ≡ 22 (mod 7) has two incongruent solutions 2x + 1 ≡ 2 (mod 7) and2x + 1 ≡ −2 (mod 7), that is, 2x ≡ 1 (mod 7) and 2x ≡ 4 (mod 7). Therefore the solutions arex ≡ 4 (mod 7) and x ≡ 2 (mod 7).

(b) Here a = 1, b = 5, c = 1, and b2 − 4ac = 21 ≡ 0 (mod 7), so all solutions are given by 2x + 5 ≡0 (mod 7), that is, 2x ≡ −5 ≡ 2 (mod 7), that is, x ≡ 1 (mod 7). So there is exactly one solution tox2 + 5x + 1 ≡ 0 (mod 7), namely, x ≡ 1 (mod 7).

(c) Here a = 1, b = 3, c = 1, and b2 − 4ac = 5. From Euler’s Criterion we have

(

5

7

)

≡ 57−12 ≡ 53 ≡ 5 · 52 ≡ 5 · 4 ≡ −1 (mod 7)

and 5 is a quadratic nonresidue of 7, so the quadratic congruence x2 + 3x + 1 ≡ 0 (mod 7) has nosolutions.

Question 11. [p 426. #2]

Using the law of quadratic reciprocity, show that if p is an odd prime, then

(

3

p

)

=

{

+1 if p ≡ ±1 (mod 12)

−1 if p ≡ ±5 (mod 12).

Solution: Let p be an odd prime, then

(i) if p = 12k + 1, then p ≡ 1 (mod 4), so that

(

3

p

)

=(p

3

)

=

(

1

3

)

= 1

(ii) if p = 12k − 1, then p ≡ 3 (mod 4), so that

(

3

p

)

= −(p

3

)

= −

(

−1

3

)

= (−1)(−1) = 1

(iii) if p = 12k + 5, then p ≡ 1 (mod 4), so that

(

3

p

)

=(p

3

)

=

(

2

3

)

= −1

(iv) if p = 12k − 5, then p ≡ 3 (mod 4), so that

(

3

p

)

= −(p

3

)

= −

(

−2

3

)

= −

(

−1

3

)

= (−1)(−1)(−1) = −1.

Therefore,(

3

p

)

=

{

+1 if p ≡ ±1 (mod 12)

−1 if p ≡ ±5 (mod 12).

Question 12. [p 426. #3]

Show that if p is an odd prime, then

(

−3

p

)

=

{

+1 if p ≡ 1 (mod 6)

−1 if p ≡ −1 (mod 6).

Solution: Let p be an odd prime, and note that

(

−3

p

)

=

(

−1

p

) (

3

p

)

=

+

(

3

p

)

if p ≡ 1 (mod 4)

−

(

3

p

)

if p ≡ 3 (mod 4).

(i) If p ≡ 1 (mod 6), then p = 6k + 1. If k is even, say k = 2m, then p = 12m + 1, so that p ≡ 1 (mod 4)and p ≡ 1 (mod 12), therefore

(

−3

p

)

=

(

3

p

)

= +1,

while if k is odd, say k = 2m + 1, then p = 12m + 7, so that p ≡ 3 (mod 4) and p ≡ −5 (mod 12),therefore

(

−3

p

)

= −

(

3

p

)

= (−1)(−1) = +1,

(ii) If p ≡ −1 (mod 6), then p = 6k−1. If k is even, say k = 2m, then p = 12m−1, so that p ≡ 3 (mod 4)and p ≡ −1 (mod 12), therefore

(

−3

p

)

= −

(

3

p

)

= −1,

while if k is odd, say k = 2m + 1, then p = 12m + 5, so that p ≡ 1 (mod 4) and p ≡ 5 (mod 12),therefore

(

−3

p

)

=

(

3

p

)

= −1.

Therefore,(

−3

p

)

=

{

+1 if p ≡ 1 (mod 6)

−1 if p ≡ −1 (mod 6).