Solutions for the Practice Questions

Solutions for the Practice Questions Question 1. Find all solutions to the congruence 13x 12 (mod 35). Also, answer the following questions about the solutions to the above congruence. Are there solutions x such that x gives a remainder of 1 when divided by 48? Are there solutions x such that x gives a remainder of 1 when divided by 49? SOLUTION. Since gcd(13, 35) = 1, we know that the congruence 13x 12 (mod 35) has exact one incongruent solution modulo 35. That is, the solutions are given by x a (mod 35), where a is an integer. The integer a is any one of the solutions. We can find the solutions by solving the linear Diophantine equation 13x 35y = 12. We have a straightforward method for solving such an equation. However, one can often find some shortcuts by being observant. One might notice that 13 3 = 39 4 (mod 35) and 13 3 3 4 3 12 (mod 35). Thus, a = 9 satisfies the congruence 13a 12 (mod 35) and the general solution to the congruence 13x 12 (mod 35) is therefore given by x 9 (mod 35). If one also wants x to give a remainder of 1 when divided by 48, then one wants x to satisfy the two congruences x 9 (mod 35) and x 1 (mod 48). Since gcd(35, 48) = 1, the Chinese Remainder Theorem implies that infinitely many such integers x will exist. Finally, if we want x to give a remainder of 1 when divided by 49, then we want x to satisfy the two congruences x 9 (mod 35) and x 1 (mod 49). Since gcd(35, 49) = 7 1, we cannot use the Chinese Remainder Theorem. In this particular situation, notice that x 9 (mod 35) implies that x 9 (mod 7). However, x 1 (mod 49) implies that x 1 (mod 7). The existence of an integer x satisfying both congruences would imply that 9 1 (mod 7). However, 9 1 (mod 7). We can therefore conclude that no such solutions x exist. Question 2. This question concerns the Diophantine equation x 2 35y 2 = 11. The purpose of this question is to prove that this Diophantine equation has no solutions. Suppose to the contrary that x = a and y = b is a solution to that equation, where a, b Z. 1

(a) Prove that gcd(a, 11) = 1 and gcd(b, 11) = 1. SOLUTION. Since 11 is a prime, we must just show that 11 a and 11 b. We are given that a 2 35b 2 = 11. It follows that a 2 35b 2 (mod 11). If 11 a, then a 0 (mod 11) and hence 35b 2 0 2 0 (mod 11). It would then follow that 11 35b 2. Thus, 11 35bb. One of the versions of Euclid s Lemma then would imply that 11 b, using the fact that 11 35. Thus, we have shown that 11 a = 11 b. On the other hand, if 11 b, then 35b 2 35 0 2 0 (mod 11) and hence a 2 0 (mod 11). Thus, 11 a 2. Euclid s Lemma would then imply that 11 a. This shows that 11 b = 11 a. In summary, we have shown that 11 a 11 b. To show that 11 a and 11 b, assume to the contrary that 11 a or 11 b. As we showed above, it then follows that 11 a and 11 b. Thus, we have a = 11k and b == 11j, where k, j Z. Since a 2 35b 2 = 11, we obtain 11 = a 2 35b 2 = (11k) 2 35(11j) 2 = 11 2 (k 2 35j 2 ) = 11 2 q, where q = k 2 35j 2. Note that q Z. Hence we obtain that 11 2 11. This is not true, and hence we have obtained a contradiction. It follows that 11 a and 11 b, as we wanted to show. (b) Show that 35b 2 a 2 (mod 11). By raising both sides of this congruence to a certain power, show that you can obtain a contradiction. SOLUTION. Since 11 a and 11 b, it follows from Fermat s Little Theorem that a 10 1 (mod 11) and b 10 1 (mod 11). As we already observed in part (a), we have 35b 2 a 2 (mod 11). Raise both sides to the 5-th power. We obtain 35 5 (b 2 ) 5 (a 2 ) 5 (mod 11). However, (b 2 ) 5 = b 10 1 (mod 11) and (a 2 ) 5 = a 10 1 (mod 11). Thus, we obtain 35 5 1 1 (mod 11). Now 35 2 (mod 11). Therefore, we have 2 5 35 5 1 (mod 11). But 2 5 = 32 10 (mod 11) and hence 2 5 1 (mod 11). We have obtained a contradiction. 2

Question 3. Find all solutions to the congruence x 5 1 (mod 11213). (Remark: You may use the fact that 11213 is a prime.) SOLUTION. Suppose that x = a is a solution. Then a 5 1 (mod 11213). Obviously, a 0 (mod 11213). This is so because a 0 (mod 11213) implies that a 5 0 5 0 (mod 11213). But 0 1 (mod 11213). It follows that 11213 a. Since 11213 is prime, we have gcd(a, 11213) = 1. We can define ord 11213 (a). Denote this quantity by e. Since a 5 1 (mod 11213). It follows that e divides 5. Thus, e = 1 or e = 5. However, 11213 is a prime, and we therefore know that e divides 11213 1 = 11212. This implies that e 5 since 5 11212. It follows that e = 1. We have a e 1 (mod 11213). Since e = 1, it follows that a 1 (mod 11213). Conversely, if a is any integer such that a 1 (mod 11213), then we also have the congruence a 5 1 (mod 11213). We have proved that x = a is a solution to the congruence x 5 1 (mod 11213) if and only if a 1 (mod 11213). Thus, the solutions to the stated congruence are described by x 1 (mod 11213). In other words, the solutions are the integers which give a remainder of 1 when divided by 11213. Question 4. Suppose that a, b, c Z and that m is a positive integer. Assume that gcd(a, m) = 1 and gcd(b, m) = 1. Assume that c ab (mod m). Prove that gcd(c 3, m) = 1. SOLUTION. By Divisibility Proposition 14, we know that gcd(ab, m) = 1. Since c ab (mod m), Congruence Property 8 tells us that gcd(c, m) = gcd(ab, m) and hence gcd(c, m) = 1. Divisibility Proposition 14 then implies that gcd(ccc, m) = 1. Therefore, we have shown that gcd(c 3, m) = 1. Question 6. Find all primes p such that ord p ( 5) = 2. Find all primes p such that ord p ( 5) = 3. Find all primes p such that ord p (3) = 8. SOLUTION. Let p be a prime. If ord p ( 5) = 2, then ( 5) 2 1 (mod p). Thus, 25 1 (mod p). Therefore, p 24. Therefore, either p = 2 or p = 3. However, notice that ( 5) 1 1 (mod 2) and ( 5) 1 1 (mod 3). We see that ord 2 ( 5) = 1 and ord 3 ( 5) = 1. We conclude that no primes p exist with the property ord p ( 5) = 2. 3

If ord p ( 5) = 3, then ( 5) 3 1 (mod p). Thus 125 1 (mod p). It follows that p divides 126 = 2 3 3 7. Thus, p = 2 or p = 3 or p = 7. However, ord 2 ( 5) = 1 and ord 3 ( 5) = 1, and hence the primes 2 and 3 don t have the stated property. Now for p = 7 divides -126 and so we do have ( 5) 3 1 (mod 7). But ( 5) 1 1 (mod 7). This suffices to show that ord 7 ( 5) = 3. The prime p = 7 is the only prime with the stated property. Finally, consider the property ord p (3) = 8. This implies that 3 8 1 (mod p), but 3 4 1 (mod p). Conversely, if 3 8 1 (mod p), but 3 4 1 (mod p)., the it follows that ord p (3) divides 8, but does not divide 4. This means that ord p (3) = 8. We have shown that ord p (3) = 8 3 8 1 (mod p), but 3 4 1 (mod p). Thus, we must find the primes p which divide 3 8 1, but do not divide 3 4 1. Now 3 8 1 = (3 4 1)(3 4 + 1) = 80 82. The primes p which divide 3 8 1 are 2, 5, and 41. The first two primes divide 3 4 1. The prime 41 does not divide 3 4 1. Thus, p = 41 is the only prime with the stated property. Question 7. This question concerns the integer n = 100000001. Prove that if p is a prime which divides n, then p 1 (mod 16). The smallest prime p satisfying p 1 (mod 16) is p = 17. Prove that 17 n. SOLUTION. Note that n = 10 8 + 1. Suppose that p is a prime and that p n. This means that 10 8 1 (mod p). Observe that p 10. To see this, note that if p 10, then 10 8 0 8 0 (mod p). Since p n, we have 10 8 1 (mod p) and therefore we would have 1 0 (mod p). That is clearly impossible. Thus, p 10 and therefore, since p is a prime, it follows that gcd(10, p) = 1. We can then define e = ord p (10). We will determine e exactly. Since p 10, we have p 2. Therefore, 1 1 (mod p). Thus, 10 8 1 1 (mod p). This shows that 10 8 1 (mod p). Therefore, e 8. We are using Congruence Theorem 5, part (a). Furthermore, the fact that 10 8 1 (mod p) implies that (10 8 ) 2 ( 1) 2 (mod p). Therefore, we have 10 16 1 (mod p). 4

Using Congruence Theorem 5(a) again, it follows that e 16. The positive divisors of 16 are 1, 2, 4, 8, and 16. Except for 16, all of those divisors of 16 actually divide 8. Since e 16, but e 8, it follows that e = 16. Finally, we use Congruence Theorem 8. That theorem asserts that e divides p 1. Therefore, 16 divides p 1. We have proved that which is what we wanted to prove. p 1 (mod 16) Now we will verify that 17 divides n. We have 10 2 = 100 2 (mod 17), 10 4 4 (mod 17), 10 8 16 1 (mod 17) and therefore 17 indeed divides 10 8 + 1 = n. Question 8. Find the remainder when 185648291 is divided by 99. We use Casting Out Nines, to conclude that 185648291 1 + 8 + 5 + 6 + 4 + 8 + 2 + 9 + 1 9 + 11 + 12 + 11 + 1 44 8 (mod 9). We have also proved a result concerning congruences modulo 11 which gives 185648291 1 9 + 2 8 + 4 6 + 5 8 + 1 18 4 (mod 11). Let n = 185648291. Then n 8 (mod 9) and n 4 (mod 11). Let e 1 = 11 5 = 55 and let e 2 = 9 5 = 45. Then e 1 1 (mod 9), e 1 0 (mod 11), e 2 0 (mod 9), e 2 1 (mod 11). Let a = 8e 1 +4e 2 = 8 55+4 45. Then we have shown in class that the two congruences n 8 (mod 9) and n 4 (mod 11) are equivalent to the single congruence n a (mod 9 11). Thus, n 8 55 + 4 45 (mod 99) Now and 8 55 8 11 5 88 5 11 5 55 44 (mod 99) 4 45 2 90 2 ( 9) 18 81 (mod 99) 5

and hence n 44 + 81 26 (mod 99). It follows that the remainder when n is divided by 99 is equal to 26. Question 9. Find the remainder when 2 19 5 8 13 11 29 18 3 600 is divided by 7. SOLUTION. By Fermat s Little Theorem, we have 2 6 1 (mod 7), 5 6 1 (mod 7), 3 6 1 (mod 7). Thus 2 19 = (2 6 ) 3 2 1 3 2 2 (mod 7), 5 8 5 6 5 2 1 25 25 4 (mod 7) and 3 600 = (3 6 ) 100 1 100 1 (mod 7). Furthermore, 13 1 (mod 7) and 29 1 (mod 7). Hence 13 11 ( 1) 11 1 (mod 7) and 29 18 1 18 1 (mod 7). Putting all of these facts together, we obtain 2 19 5 8 13 11 29 18 3 600 2 4 ( 1) 1 1 9 5 (mod 7) and therefore the given integer gives a remainder of 5 when divided by 7. 6