exam2.sols

8/12/2019 Exam2.Sols

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Math 580, Exam 2. 10/21/09. Name:

Read problems carefully. Show all work.

No notes, calculator, or text.

The exam is approximately 15 percent of the total grade.

There are 100 points total. Partial credit may be given.


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1. (25 points)

(a) (10 points) Find all incongruent solutions mod44to the linear congruence

19x 8 (mod 44).

Solution: gcd(19, 44) = 1|8 = there is one solution mod44.19x 8 (mod 44) = k Zwith19x 44k = 8. We solve19x 44k = 1asfollows.

44 = 2 19 + 6

19 = 3 6 + 1

We have:

1 = 19 3 6 = 19 3 (44 2 19) = 19 7 44 3

Multiplying by8gives19 56 44 21 = 7,

from which it follows that x 56 12 (mod 44).

(b) (8 points) How many incongruent solutions mod51does the linear congruence

17x 34 (mod 51)have? Briefly explain. I am not asking you to find the solutions(if any exist).

Solution: gcd(17, 51) = 17| 34 = there are17solutions modulo51.

(c) (7 points) How many incongruent solutions mod68does the linear congruence52x 14 (mod 68)have? Briefly explain. I am not asking you to find the solutions(if any exist).

Solution: We have

68 = 1 52 + 1652 = 3 16 + 4

16 = 4 4 = gcd(52, 68) = 4.

Since4 14, there are no solutions.

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2. (20 points) Solve the following system of linear congruences.

x 3 (mod 4)

x 2 (mod 7)

x 7 (mod 9).

You may leave your answer as asum of products (unsimplified).

Solution:

7 9 x1 1 (mod 4) 63x1 3x1 1 (mod 4) x1 3 (mod 4).

4 9 x2 1 (mod 7) 36x2 x2 1 (mod 7).

4 7 x3 1 (mod 9) 28x3 x3 1 (mod 9).

Now, the solution is

x (7 93)3+(491)2+(471)7 567+72+196 835 79 (mod 252 = 479).

Note: You can check your answer.

79 3 (mod 4)

79 2 (mod 7)

79 7 mod 9.

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3. (20 points)

(a) (10 points) Letnbe a positive integer. Show that17 |3 52n+1 + 23n+1.

Solution: We have

3 52n+1 + 23n+1 = 3 25n 5 + 8n 2

15 8n + 2 8n

17 8n 0 (mod 17).

(b) (10 points) UseFermats Little Theoremto compute3226 (mod 23).

Solution: Sincegcd(3, 23) = 1, Fermats Little Theorem =

322 1 (mod 23).

We also note that226 = 10 22 + 6

. It follows that

3226 = (322)10 36 36 (mod 23).

But also, we have

33 = 27 4 (mod 23) = 36 = (33)2 42 16 (mod 23).

Hence, we have3226 16 (mod 23).

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4. (20 points)

(a) (5 points) One of the following statements is true. Circle the true statement. Cite a

theorem from class to justify your answer.

There are infinitely many primes of the form15n+ 6.

There are infinitely many primes of the form20n+ 9.

Solution: gcd(15, 6) = 3;gcd(20, 9) = 1. Hence, Dirichlets Theorem = thereare infinitely many primes of the form20n+ 9.

(b) (5 points) State Bertrands Postulate. It is sometimes called Bertrands Conjecture

(but it is actually a theorem).

Solution: For alln 2, there is a primep withn < p 1 be an integer. Use Bertrands Postulate to show that n!is not a

perfect square. (In other words,n!=m2 for any integerm.)

Solution: Note thatn!is square the power of all primespdividingn!is even.

n even = prime p with n/2 < p < n. The power of this prime in thefactorization ofn!is one (odd). Hence, n!is not square.

nodd = primepwith(n + 1)/2< p < n + 1The power of this prime in thefactorization ofn!is one (odd). Hence, n!is not square.

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5. (15 points) Suppose thatp and qare distinct odd primes, that a Z, and that gcd(a,pq) = 1.Prove that

a(p1)(q1)+1 a (mod pq).

Solution: gcd(a,pq) = 1 = pa, so by Fermats Little Theorem, we have

ap1 1 (mod p) = a(p1)(q1)+1 (ap1)q1 a 1 a a (mod p).

Similarly,gcd(a,pq) = 1 = qa; by the same reasoning, we find that a(p1)(q1)+1 a(modq). Now, sincep and qare distinct primes, the result follows (by, for example, theChinese Remainder Theorem).

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