Practice Final Solutions

Transcription

1 Practice Final Solutions 1. True or false: (a) If a is a sum of three squares, and b is a sum of three squares, then so is ab. False: Consider a 14, b 2. (b) No number of the form 4 m (8n + 7) can be written as a sum of two squares. True: Since it cannot be written as a sum of three squares, it cannot be written as a sum of two squares. (c) A number can be both a quadratic residue modulo an odd rime, and a rimitive root modulo. False: If a is a quadratic residue mod then by Euler s criterion a ( 1)/2 1 and so the order of a is not 1. (d) A erfect number must have at least four divisors. True: If n > 1 is erfect then σ(n) 2n. If it has just two divisors then σ(n) n + 1 < 2n. Hence it must have a third divisor d and thus also a fourth divisor n/d (which may be d and then you can argue why there must then be yet another divisor). (e) An infinite simle continued fraction never converges to a rational number. True: roved in class. 2. Let be an odd rime. Find a formula for the Legendre symbol ( ) 2 SOLUTION: One has ( ) 2 ( 1 ) ( ) 2 Let us find a formula for when this is equal to 1. The two cases are and ( 1) ( 1) 1. The first is the case if 1 mod 4 and ±1 mod 8. This means that one needs 1 mod 8. The second is the case if 3 mod 4 and 3 or mod 8. This means that 3 mod 8. Hence ( ) 2 1 1

2 if and only if 1 or 3 mod Let 1 mod 4 be a rime and a a quadratic residue mod. Decide with justification if then automatically a is quadratic residue mod. SOLUTION: Since a a mod it suffices to decide if a is a quadratic residue. But ( ) ( ) ( ) a 1 a ( 1) ( 1)/2 1 1 since a is a quadratic residue and 1 mod (a) Calculate φ(7!). SOLUTION: Note that 7! Hence φ(7!) φ(2 4 )φ(3 2 )φ()φ(7) (b) Suose and q are twin rimes, i.e. q + 2. Show that φ(q) φ() + 2 SOLUTION: Since and q are rimes one has φ(q) q ( 1) + 2 φ() + 2 (c) Suose n 2 2k Find τ(n) and σ(n). SOLUTION: τ(n) 2 k + 1 σ(n) 2 2k Find the 8th convergent of 2 As we saw in class, 2 [1, 2]. We find 7 /q 7 (if you found 8 /q 8, that s ok). 0 1, , , , , , , q 0 1, q 1 2 2, q , q , q , q , q , q Hence the eighth convergent, C 7 is 77/408. 2

3 6. Suose is rime and n 2 and a 2 1 mod n. Show that ord n a 2 if and only if a 1 mod n. SOLUTION: Since a 2 1 mod n it follows that the order of a divides 2. It is hence one of 1,, 2 since is a rime. It is then clear that the order is 2 recisely when a 1 mod n. 7. Let n 1. Find SOLUTION: Note that gcd(n, 2n 2 + 1) (2n 2 + 1) 2n n 1 It follows that the gcd is Find a number λ between 200 and 20 such that for every n φ(λ) there exists an integer A such that ord λ (A) n. SOLUTION: 233 is a rime number, so there is a rimitive root r mod 233. So, if n φ(233) 232, we have that the order of r 232/n is recisely n mod 233, since it is ord 233 (r)/(ord 233 (r), 232/n) 232/(232/n) n. 9. (a) State the Mobius inversion formula. (b) Show that for all n 1 one has τ(d)µ(n/d) 1 d n SOLUTION: Let By Mobius inversion one has But one also has Hence and hence and hence f τ µ τ f 1 τ 1 1 (f 1) µ (1 1) µ f δ 1 δ f 1 and this imlies the desired result. 3

4 10. (a) Find all rimitive roots modulo 23. First, note that is a rimitive root mod 23, since its order mod 23 must divide φ(23) 22, and so it must be 2, 11, or 22. We have 2 2 and 1 1 ( 2 ) (mod 23), and so ord 23 () 22. Then we have that the set of rimitive roots is, 3,, 7, 9, 13, 1, 17, 19, 21, since all of these owers have gcd 1 with ord 23 () 22. (b) How many rimitive roots are there modulo 171? SOLUTION: 171 is 9 19, and by the rimitive root theorem there are no rimitive roots modulo a number of this form (since it is not a ower of a rime, or twice the ower of a rime). (c) How many rimitive roots are there modulo 173? SOLUTION: 173 is rime, so there are φ(φ(173)) φ(172) φ(4 43) rimitive roots (mod 1) How many rimitive roots are there modulo ? SOLUTION: None: by the Primitive Root Theorem, only modulo numbers of the form 1, 2, 4, m, and 2 m where is an odd rime and m 1 is an integer can one have a rimitive root. 12. Find the order of 12 modulo 3. SOLUTION: We have that this order must divide φ(3) 24, and cannot be 24 since there are no rimitive roots modulo 3. Hence we check that (mod 3), (mod 3), (mod 3), (mod 3). The only remaining ossible order is 12 is 12, so this is the order. ord 3 (12) Write down the continued fraction exansion for 29. Find its first five convergents. SOLUTION: 29, so the first term of the exansion is. The next term is The next term is /( 2)

5 The next term is /( 1) The next term is /( 1) The next term is /( 2) The next term is from before. Clearly, from now on the terms will reeat the eriod 2, 1, 1, 2, 10, so the continued fraction exansion is 29 [, 2, 1, 1, 2, 10]. The first five convergents are which are, 11/2, 16/3, 27/, 70/13., [, 2], [, 2, 1], [, 2, 1, 1], [, 2, 1, 1, 2], 14. Which quadratic irrational does the continued fraction [4, 2, 1] corresond to? SOLUTION: First we note that if β [2, 1], then β , β so, so 1 2 β + 1 β + 1 β 2 + β 2β β 3β + 2, and so β 2 2β 2 0, meaning β or β

6 It cannot be the latter, as it is ositive, so β Now, we have [4, 2, 1] β For which ositive integers a is (a+ )/3 exressed as an eventually eriodic continued fraction? A eriodic continued fraction? SOLUTION: Since this is a quadratic irrational for all integers a, it has an eventually eriodic continued fraction for all integers a. In order for the continued fraction to be eriodic, we must furthermore have that (a + )/3 > 1, meaning that a 1. Finally, we must have that 1 < (a )/3 < 0, and so a 2. Thus a may be 1 or 2 for this to be a eriodic continued fraction. 16. Find two continued fraction exansions for 13. Are there others? Why or why not? SOLUTION: Run the Euclidean algorithm on (13, ) and get Thus one continued fraction exansion of 13/ is [2, 1, 1, 2], and another is [2, 1, 1, 1, 1], since 1/2 1/(1 + 1/1). These are the only two exansions, since rational numbers always have exactly two exansions, as roven on the homework. 17. Show that is a convergent of This isn t a great ractice roblem, because its best done with a calculator. However, it hels us to recall an imortant result. If you comute , this is < 1 2(2911) , and so this must be a convergent by a theorem shown in class. 18. Which of the following can be written as a sum of two squares? A sum of three squares? Four squares? (a) (b) (c)

7 (d) 12! (e) A number of the form 2 + 2, where is a rime. SOLUTION: For the first three of these, one first writes down the rime factorizations: they are , , All of these contain rimes in the rime factorization which are 3 mod 4 but not taken to an even ower, so none are sums of two squares. Also, 3 is a rime divisor of 12!, but it divides 12! exactly u to the fifth ower, which is not even, and hence 12! is also not a sum of two squares. Finally, for the last examle, if is even, which is not a sum of two squares, and (mod 4) if is odd, so this is never a sum of two squares. Now we check for sums of three squares. The only numbers not exressible as sums of three squares are of the form 4 m (8n + 7). None of the first three examles are divisible by a ower of 4, so we just check them mod 8. The first examle is even mod 8, the second examle is 3 mod 8, so it can be written as a sum of three squares. The third examle is again even mod 8, and so it also can be written as a sum of three squares. The fourth examle is , where is 7 mod 8, so it cannot be written as a sum of three squares. The last examle can always be written as All of these can be written as a sum of four squares, since all integers can be. 19. Suose x Z >0 can be written as a sum of two squares. What is the necessary and sufficient condition on y Z >0 for xy to be exressible as a sum of two squares? SOLUTION: The necessary and sufficient condition is that y is a sum of two squares. It is sufficient because the roduct of two integers that are sums of two squares is also a sum of two squares. To see that it is necessary, suose y cannot be written as a sum of two squares, but x can. FIrst of all, this means that there is a rime that is 3 mod 4 and such that the highest ower of dividing y is an odd ower. Second of all, whenever is such a rime, the highest ower of dividing x is even (maybe 0). So the highest ower of dividing xy is odd, and so xy cannot be written as a sum of two squares. 20. Show that the area of any right triangle with all integer sides is divisible by 6. SOLUTION: Given the arametrization of rimitive Pythagorean triles (x, y, z), we have that x m 2 n 2, y 2mn, z m + n 2 for some integers m, n which are relatively rime, and such that exactly one of them is even. Since one of them is even, y is divisible by 4. Suose m is even, n is odd, and neither is divisible by 3. Then both m 2 and n 2 have to be 1 mod 3, and so x is divisible by 3, and the area, xy/2, is divisible by 12/2 6. If one of m, n is divisible by 3, then y is divisible by 12 and so the area is divisible by 6. 7

8 21. Which rimes can be the hyotenuse (i.e. largest integer) in a rimitive Pythagorean trile? Justify your answer. SOLUTION: In a rimitive Pythagorean trile, the hyotenuse is always a sum of two squares, m 2 + n 2. Hence any rime that is 1 (mod 4) can be the hyotenuse of a PPT, and no others (2 is not a hyotenuse for a PPT). Note that if the hyotenuse is rime, then the trile must be rimitive unless one of the legs is 0. This would mean either m or n is 0 (not ossible since is not a square) or that m n, which would mean 2m 2 which is true for no rime excet 2. 8