MATH 3240Q Introduction to Number Theory Homework 7

Transcription

1 As long as algebra and geometry have been searated, their rogress have been slow and their uses limited; but when these two sciences have been united, they have lent each mutual forces, and have marched together towards erfection. Joseh Louis Lagrange. Question 1. The following is a table of owers of 2 modulo 13: x x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 x 11 x Without finding them, how many rimitive roots are there in Z/13Z? 2. Find all rimitive roots of Use the table to find all quadratic residues modulo From the given table we clearly see that 2 is a rimitive root. Then, there are φ(φ(13 φ(12 φ(4φ(3 4 rimitive roots. 2. The rimitive roots coincide with those owers 2 n with n relatively rime to 12, i.e. n 1, 5, 7, 11 and the rimitive roots are 2, 6, 7 and Since 2 is a rimitive root, the quadratic residues are those 2 n with n even. Hence, the quadratic residues are 4, 3, 12, 9, 10, 1 (and there are of them. Question 2. The fraction 3/14 has decimal exansion so we say that the eriod has length 6 and the osition of the first digit in the eriod (within the exansion of the fractional art is the second digit of the exansion. Without actually doing the long division, find out the length of the eriod, and the osition of the first digit in the eriod, for each of these fractions: 1. 2/ / / / / In the solutions for this roblem I am referring to the notation of Section 8.9 and Theorem in the book. 1. 2/ Here so M 1 and the reeating art of the decimal exansion starts with the second digit, and the order of 10 modulo 7 is 6, i.e., ord 7 (10 6. Hence, the length of the eriod is 6.

2 2. 11/ Here so M 4 and the reeating art of the decimal exansion starts with the M + 1 5th digit, and the order of 10 modulo 13 is 6, i.e., ord 13 (10 6. Hence, the length of the eriod is / Here 17 is rime so M 0 and the reeating art of the decimal exansion starts with the M + 1 1st digit, and the order of 10 modulo 17 is 16, i.e., ord 17 ( Hence, the length of the eriod is / Here 97 is rime so M 0 and the reeating art of the decimal exansion starts with the M + 1 1st digit, and the order of 10 modulo 97 is 96, i.e., ord 97 ( Hence, the length of the eriod is / Here where 97 is rime so M 3 and the reeating art of the decimal exansion starts with the M + 1 4th digit, and the order of 10 modulo 97 is 96, i.e., ord 97 ( Hence, the length of the eriod is 96. Question 3. Is 45 a quadratic residue modulo? Notice that , so ( 45 ( Now, we use quadratic recirocity. Since 5 1 mod 4 then: where we have used quadratic recirocity and then the fact that 2 mod 5. Therefore, 45 is not a quadratic residue modulo. Here is an alternative way: 45 2 mod. Thus: ( 45 ( 2 ( 2 Since 7 mod 8, 2 is a QR and since 3 mod 4, 1 is not a square modulo, and so: ( Question 4. Is 13 a square modulo? By the roerties of the Legendre symbol: ( 13 1 ( 13

3 where we have used the fact that 1 mod 4 and, thus, 1 is a quadratic residue modulo. Now we shall use Quadratic Recirocity. Since 13 1 mod 4 and 11 mod 13: and, again by Quadratic Recirocity and 13 1 mod 4: Finally, since 11 3 mod 8, the number 2 is not a square modulo 11 and so: Hence, 13 is not a quadratic residue modulo. Question 5. Is 14 a quadratic residue modulo 65? We need to check whether x 2 14 mod 65 has solutions. By the Chinese Remainder theorem, and since , this equation has solutions if and only if the system: { x mod 5 x mod 13 has solutions. But, clearly, these systems have solutions. Therefore, the original system has a solution as well. For examle, a solution of the system x 2 mod 5 and x 1 mod 13 will work. Thus x 27 mod 65 is a solution of x 2 14 mod 65. Question 6. Find the following values of the Legendre symbol: ( ,,, The numbers 113, 127, 131, 1 are rimes. For this, use the law of quadratic recirocity. In the last one, first factor , 1, 1, ( ( ( ( Question 7. Find the value of the following Legendre symbol: rime but 4699 is not! ( Note that is

4 The number 4699 factors as 127, and notice that they are both 1 mod 4, thus, by the roerties of the Legendre symbol and quadratic recirocity: ( 4699 ( 4 ( 4 1 because 3 mod 4. Another way, using Quadratic Recirocity: ( ( ( ( ( ( Question 8. Find all solutions (if any of the equations: x x mod 1, x 2 + 5x mod, x 2 + 5x mod Use the quadratic formula. First one needs to calculate the discriminants of the equations: , , Since: ( 113 1, 1 ( 13 1, ( 3 1 the first equations have no solutions and the third one does. In fact mod and the solutions are 8 and 24 modulo and (x 8(x 24 x 2 + 5x + 7 mod. Question 9. Prove that the equation x 2 1y has no integer solutions. Suose that (x, y is a solution in integers and reduce the equation modulo 1, to obtain x mod 1. But in roblem 1 we saw that (113/1 1, so 113 is not a quadratic residue modulo 1 and x mod 1 cannot have solutions. Contradiction. Question 10. For what rimes is 5 a quadratic residue? For what rimes is 10 a quadratic residue? Are there infinitely many rimes such that 10 is a quadratic residue modulo? Hint: Use Quadratic Recirocity! Use quadratic recirocity: ( 5 ( 5 ( 5.

5 Thus, 5 is a QR if 1 is a QR and ±1 mod 5, or 1 is not a QR and 2, 3 mod 5. Notice that 1 is a QR if and only if 1 mod 4, so one can exress all this in terms of congruences modulo 20 (do it!. Similarly, one can treat the question about 10 (in this case the answer will be modulo 40. And there are infinitely many such rimes (use Dirichlet s theorem. Question 11. Are there two odd rimes, q such that q, q 3 mod 4 and such that is a quadratic residue modulo q and q is a quadratic residue modulo? What is the smallest odd rime q such that 3 is a quadratic residue modulo q and q is a quadratic residue modulo 3? The answer to the first question is no (use quadratic recirocity. For the second question, we need to find such that is a QR modulo 3 and 3 is a QR modulo. Hence, must be 1 modulo 4, and must be 1 modulo 3. By the Chinese Remainder Theorem, 1 mod 12. The first such rime is 13. Question 12. Use induction to show that, for all n, there exists a set of n distinct odd rimes { 1,..., n } such that ( i 1 j for all 1 i, j n with i j, i.e. every rime in the list is a quadratic residue modulo any other rime in the list. We will show that this can be done with all i 1 mod 4. For n 1 it is clear. Now let { 1,..., n 1 } be a set of such rimes, i.e. i 1 mod 4 and ( i / j 1 for i j. Then, by Dirichlet s theorem on rimes in arithmetic rogressions there are infinitely many rimes such that 1 mod n 1. Let n be one of them. In articular, n 1 mod 4 and n 1 mod i for all 1 i n q (by the Chinese Remainder Theorem. Hence: i n 1 1 for all 1 i n 1 as desired. n i i Question 13. Suose that and q are twin rimes. Is it ossible that 2 is a quadratic residue for both and q? Is 2 necessarily a quadratic residue of or q? Find twin rimes and q such that 2 is a quadratic residue modulo but not modulo q. Suose and q are twin, with q + 2. The number 2 is a QR modulo a rime t if and only if t ±1mod8. Suose 2 is a QR modulo, then ±1 mod 8 but then or 1 modulo 8. Hence, if 2 is a QR for q + 2, then cannot be 1 modulo 8, but there is no roblem if 1 mod 8 and q 1 mod 8. For examle, 71 and 73 are both rimes (twin and 2 is a QR for both of them.

6 If and q + 2 are twin rimes but 3 mod 8 then q 5 mod 8 and for neither of them 2 is a QR. For examle, 11 and 13. By our first remarks, if 2 is a QR for but not for q + 2 then 1 mod 8. For examle, 17 and 19. Question 14. The number 4003 is rime. 1. The number 2 has exact order 2001 modulo Find a rimitive root modulo. 2. The number 285 has exact order 87 modulo, and the number 2163 has exact order 46 modulo. Find another rimitive root mod. 3. Given that mod 4003 and mod 4003, rove that 2 is also a rimitive root modulo Let us assume (as they tell us to do that 2 has exact order 2001 modulo A number is a rimitive root modulo 4003 if its order is exactly Notice that has order 2 modulo Since (2001, 2 1, then the order of ( 1 2 is equal to Therefore, mod 4003 is a rimitive root. 2. Suose ord( and ord( Notice that and and Since (87, 46 1 then ord( Hence, mod 4003 is also a rimitive root. 3. We are given mod 4003 and mod Recall the formula: Thus: and Hence: ord(a n ord(a (ord(a, n. 46 ord(2163 ord(2 87 ord(2 (ord(2, ord(285 ord(2 46 ord(2 (ord(2, 46. ord(2 46 (ord(2, (ord(2, 46. Consequently, the order of 2 is divisible by 46 and by 87. Since these two numbers are relatively rime, ord(2 is divisible by Since the order of any number modulo divides 1, we conclude that ord( and 2 is a rimitive root.