MTH 7 Number Theory Quiz 10 (Some roblems that might be on the quiz and some solutions.) 1. Euler φ-function. Desribe all integers n such that: (a) φ(n) = Solution: n = 4,, 6 since φ( ) = ( 1) =, φ() = 1 =, φ(6) = φ()φ() = 1 = (b) φ(n) = 4 Solution: n = 8, 1,, 10 since φ( ) = () = 4, φ( ) = = 4, φ() = = 4, φ( ) = 1 ( 1) = 4. (c) φ(n) = 8 Solution: n = 16, 4, 0, 1, 0 (d) φ(n) = 16 Solution: n = 17, 4, 60, 40, 48,, which is the same as: n = 17, 17,,, 4,, (e) φ(n) = (f) φ(n) =. Solution: No such n exists by (m,n,o) arts of this roblem. (g) φ(n) = 9. Solution: No such n exists by (m,n,o) arts of this roblem. (h) φ(n) = 6 Solution: n = 7, 7, 9, 9 (i) φ(n) =. Solution: No such n exists by (m,n,o) arts of this roblem. (j) φ(n) = 10 Solution: n =, (k) φ(n) = 0 Solution: n =,,,, (l) φ(n) = 40 Solution: n = 41, 8,, 0, 1, 88, 7, 10, 100 n = 41, 41,,,,,,, (m) Prove that φ() is even integer for all odd rimes. Solution: If is rime, then φ() = 1. If is odd, then 1 is even. Hence φ() is even. (n) Prove that φ( k ) is even integer for all odd rimes. Solution: If is rime, then φ( k ) = k ( 1). So, if is odd, then 1 is even, hence φ( k ) is even. 1
(o) Prove that φ(n) is even integer for all n >. Solution: If a, b are relatively rime integers we have φ(ab) = φ(a) φ(b). Case 1. There is an odd rime q, such that q k aears in the rime factorization of n. Then (q 1) φ(q k ) φ(n). Hence φ(n) is even. Case. n = k. From the assumtion n >, we have that k. So k = φ( k ) () Prove that the only integers n, such that φ(n) = 0 are n = 1, 6. (q) Prove that there are no integers n such that φ(n) = 0.. Roots of olynomials (mod ): Theorem: Let be a rime. If n is a ositive integer, such that n (), then x n 0 (mod) has exactly n solutions with 0 x <. (a) Show that x 1 0(mod 1) has non-congruent roots. Solution: Since 0, we may aly the above theorem. So there are solutions with 0 x < 0. Since all integers {0, 1,..., 0} are non-congruent modulo 1, these solutions are non-congruent roots modulo 1. (b) Show that x 1 0(mod ) does not have non-congruent roots. It has only one root. Solution: 0 0(mod ), 1 1(mod ), (mod ), (mod ), 4 4(mod ). Therefore x = 1 is the only root for x 1 0(mod ) with 0 x < 4. (c) Show that x + x + 1 has roots (mod 1). Solution: x 1 0(mod 1) has non-congruent roots, since 0. x 1 0(mod 1) has 1 (non-congruent) root since 1 0. x 1 = (x 1)(x + x + 1) 0(mod 1) has non-congruent roots, x 1 has one root, imlies that (x + x + 1) 0(mod 1) has roots. (d) Show that x + x + 1 0(mod ) does not have any roots. Solution: 0 + 0 + 1 1(mod ), 1 + 1 + 1 (mod ), + + 1 (mod ), + + 1 (mod ), 4 + 4 + 1 1(mod ). Therefore no x is a root for x + x + 1 0(mod ). (e) Show that x 1 has non-congruent roots (mod 1). (f) Show that x 1 has non-congruent roots (mod ). (g) Show that x 1 has non-congruent roots (mod 101). (h) Show that x 0 1 has 0 non-congruent roots (mod 101). Solution: Since 0 (101 1), we may aly the above theorem and get 0 non-congruent solutions. (i) Show that x 4 + x + x + x + 1 has 4 roots (mod 1). Solution: x 1 0(mod 1) has non-congruent roots, since 0. x 1 0(mod 1) has 1 (non-congruent) root since 1 0. x 1 = (x 1)(x 4 + x + x + x + 1) 0(mod 1) has non-congruent roots, x 1 has one root, imlies that (x 4 + x + x + x + 1) 0(mod 1) has 4 roots.
(j) Show that x 4 + x + x + x + 1 has 4 roots (mod ). (k) Show that x 4 + x + x + x + 1 has 4 roots (mod 101). (l) Show that x 10 1 has 10 roots (mod ). (m) Show that x + 1 has roots (mod ). Solution: x 10 1 0(mod ) has 10 non-congruent roots, since 10 ( 1). x 1 0(mod ) has (non-congruent) roots since 10. x 10 1 = (x 1)(x + 1) 0(mod ) has 10 non-congruent roots, x 1 has roots, imlies that (x + 1) 0(mod ) has roots. (n) Show that x 8 + x 6 + x 4 + x + 1 has 8 roots (mod ). Solution: Use the following factorization, and arguments similar to the revious roblems. x 10 1 = (x ) 1 = (x 1)((x ) 4 + (x ) + (x ) + x + 1) 0(mod ). Most of you did not state the Rabin-Miller test correctly. So here are some roblems to clarify how the test works (or doesn t). (a) State Rabin-Miller test. (b) Use a = as a Rabin-Miller witness that 1 is comosite. (c) Use a = as a Rabin-Miller witness that 1 is comosite. (d) Use a = 4 as a Rabin-Miller witness that 1 is comosite. (e) Use a = as a Rabin-Miller witness that 7 is comosite. (f) Use a = as a Rabin-Miller witness that 7 is comosite. (g) Use a = as a Rabin-Miller witness that 81 is comosite. (h) Can you use a = as a Rabin-Miller witness that 0 is comosite? Exlain. (i) Can you use a = as a Rabin-Miller witness that 0 is comosite? Exlain. (j) Use a = in Rabin-Miller test to check if 7 is comosite. (k) Use a = in Rabin-Miller test to check if 7 is comosite. (l) What haens if you use a = in Rabin-Miller test to check if 101 is comosite? (m) What haens if you use a = in Rabin-Miller test to check if 101 is comosite? a 4. Legendre symbol:. Find the following: (a) Solution: = since = when the rime (mod 4).
(b) = Solution: = 1 since = 1 when the rime 1(mod 4). 1 1 (c) = Solution: = 1 since = 1 when the rime 1, 7(mod 8). 17 17 (d) = Solution: = since = when the rime, (mod 8) 1 1 (e) 101 (f) = Solution: = = ()() = 1, since the rime (mod 4) and (mod 8) (g) (h) (i) (j) (k) (l) = Solution: 1.. 6 = 1 9 = 10 = 10 00 = 10 1 = since (mod 4) and 1 (mod 4). 1 1 1 = since 1 1(mod ). 1 = 1 since 1 is quadratic residue (mod), which is true since 1 1 (mod). 4
. For each of the following congruences determine if it has a solution: (a) x (mod 987). (987 is rime.) Solution: The congruence x (mod 987) has a solution iff is a square modulo 987 (which is rime) iff is quadratic residue modulo 987 iff = 1 iff 987 987 1(mod4). Since 987 (mod4) it follows that is not a square modulo rime 987 and therefore there is no solution to x (mod 987). (b) x (mod 987). (987 is rime.) (c) x (mod 987). (987 is rime.) (d) x (mod 987). (987 is rime.) (e) x 10(mod 987). (987 is rime.) 6. For each of the following integers determine if it is a quadratic residue modulo, i.e. QR : (a) (mod 887). (887 is rime.) (b) (mod 887). (887 is rime.) (c) (mod 16). (16 is rime.) (d) (mod 167). (167 is rime.) (e) 10(mod 8). (8 is rime.) Solution: 10 is QR 8 iff the Legendre symbol is =1. 10 8 Comute = = () = () = () = 8 8 8 () = ()() = 1. Therefore 10 is QR 8.