HOMEWORK # 4 MARIA SIMBIRSKY SANDY ROGERS MATTHEW WELSH 1. Section 2.1, Problems 5, 8, 28, and 48 Problem. 2.1.5 Write a single congruence that is equivalent to the air of congruences x 1 mod 4 and x 2 mod 3. Solution. x 1 mod 4 and x 2 mod 3 if and only if x 5 mod 12. Proof. Note: 3, 4 1, so the Chinese Remainder Theorem alies. We have φ : Z/12Z Z/3Z Z/4Z, where the ma φ is given by φx x mod 3, x mod 4. We need to find φ 1 x. Since 4 1 mod 3, we note that 2 4 2 mod 3 and 2 4 mod 4. Similarly, 3 3 1 mod 4 and 3 3 mod 3. Therefore, if we let x 2 4 + 1 3 3, we have x 2 mod 3 and x 1 mod 4. Reducing x modulo 12, we have, x 24 + 33 17 5 mod 12, so φ5 2, 1 as desired. Problem. 2.1.8 Prove that any number that is a square must have one of the following for its units digit:, 1, 4, 5, 6, 9. Proof. By Theorem 2.2 in the boo, we now a b mod 1 imlies a 2 b 2 mod 1. Every x Z is equivalent mod 1 to some a {, 1, 2, 3, 4, 5, 6, 7, 8, 9}. It follows that x 2 a 2 mod 1. The set formed by a 2 is: {, 1, 4, 9, 6, 5, 6, 9, 4, 1} {, 1, 4, 5, 6, 9}. Note that we only have to chec,..., 5 since a 2 1 a 2 mod 1. Problem. 2.1.28 What is the last digit in the ordinary decimal reresentation of 3 4? Solution. The answer is 1 because 3 4 1 mod 1. Proof. Theorem 2.2 states that a b mod m imlies a n b n mod m. Since 3 4 81 1 mod 1, It follows 3 4 3 4 1 1 1 1 mod 1. Problem. 2.1.48 If r 1, r 2,..., r and r 1, r 2,..., r are any two comlete residue systems modulo a rime > 2, rove that the set r 1 r 1, r 2 r 2,..., r r cannot be a comlete residue system modulo. Proof. Any comlete residue system modulo is a ermutation of the equivalence classes, 1, 2,..., 1 in Z/Z. Let i and j be the indices such that r i r j mod. If i j, then the elements r i r i and r j r j are both equivalent to mod, so they cannot be art of a comlete residue system mod. Therefore, we can assume i j. After removing r i and 1
r i from these lists, we thus see that the remaining elements, which we relabel r 1,..., r 1 and r 1,..., r 1, are comlete residue systems for Z/Z, i.e. they reresent the nonzero classes mod in some ermuted order. Therefore, by Wilson s Theorem, Hence, r 1 r 2 r 1 1! 1 mod r 1r 2 r 1 1! 1 mod r 1 r 1 r 2 r 2 r 1 r 1 1 1 1 mod If r 1 r 1 r 1 r 1 was a comlete residue system for Z/Z, we would have r 1 r 1 r 1 r 1 1 mod. But 1 1 mod since > 2. Therefore, r 1 r 1 r 1 r 1 is not a comlete residue system for Z/Z, and hence the original r 1 r 1 r r is not a comlete residue system mod as desired. 2. Section 2.2, Problems 8, 9, and 14 Problem. 2.2.8 Show that if is an odd rime then the congruence x 2 1 mod has only the two solutions x 1 and x 1 mod. Proof. x 2 1 mod x 2 1 x 1x + 1 Case 1:, x 1 > 1. Then x 1 since the only divisors of are owers of. Since > 2, does not divide x + 1, so, x + 1 1. By the fundamental theorem of arithmetic, we see that x 1x + 1 x 1. Therefore x 1 mod. Case 2:, x 1 1. Then by the fundamental theorem of arithmetic, we see that x 1x + 1 x + 1. Therefore x 1 mod. Problem. 2.2.9 Show that the congruence x 2 1 mod 2 has one solution when 1, two solutions when 2, and recisely four solutions 1, 2 1 1, 2 1 + 1, 1 when 3. Proof. When 1, x or 1 mod 2. Squaring both of these shows that x 2 1 if and only if x 1 mod 2. Similarly, when 2, x, 1, 2, or 3 mod 4. Squaring shows that x 2 1 imlies that x 1 or x 3 mod 4. Now consider the case when 3. x 2 1 mod 2 imlies, by definition, that 2 x 2 1, or equivalently 2 x 1x + 1. There are three cases to consider. Case 1: 2, x+1 1. Then 2 x 1 by the fundamental theorem of arithmetic, so x 1 mod 2. Case 2: 2, x 1 1. Then 2 x + 1 by FTA, so x 1 mod 2. Case 3: Both 2, x 1 > 1 and 2, x+1 > 1. Since the only divisors of 2 are owers of 2, this imlies that 2 x+1 and 2 x 1. This, along with the fact that 2 x+1x 1, imlies that 2 2 x+1 x 1 x+1. Note that > 2 so this is still well-defined. Since x 1 + 1, 2 2 2 2 either 2 2, x+1 1 or 2 2 2, x 1 x 1 1, deending on whether is even or odd. The 2 2 fundamental theorem of arithmetic imlies that either 2 2 x+1 or 2 2 x 1. This, in turn, 2 2 imlies that 2 1 x + 1 or 2 1 x 1. This leaves us with four ossible congruences of x modulo 2, two of which have already been considered secifically x 1 or 1 mod 2. The two new ossibilities are x 2 1 1 and x 2 1 + 1 mod 2. 2
In conclusion, x 2 1 mod 2 has four solutions when 3. These are 1, 2 1 1, 2 1 + 1, and 1. Problem. 2.2.14 Show that mod for < < and use this to show that a + b a + b mod. Proof. First note that Indeed, we calculate Now 1 1.!!! 1! 1! 1 1! 1 1 Z Let m β where m, 1 and β <. Then we have 1 m β. 1 1 1 The fundamental theorem of arithmetic imlies that 1 1 m β 1 1 1 1 1 So 1 1 1 Z 1 1 1 1. mod. This roves the first art of the roblem. Now we show that a+b a +b a + b a b a b + + a b + 1 1 a b mod Notice that 1 and 1 so a b + + a b b + a a b + mod. a b mod. 3
3. Section 2.3, Problems 15, 2, and 47 Problem. 2.3.15 Solve the congruence x 3 + 4x + 8 mod 15. Proof. Let fx x 3 + 4x + 8. x must be in some equivalence class modulo 5, where Z/5Z {[ 2], [ 1], [], [1], [2]}. Trying the values x, ±1, ±2, we find that fx mod 5 has no solutions. Since 5 15, it follows that fx mod 15 has no solutions. Problem. 2.3.2 Let m 1 and m 2 be arbitrary ositive integers, and let a 1 and a 2 be arbitrary integers. Show that there is a simultaneous solution of the congruences x a 1 mod m 1 and x a 2 mod m 2 if and only if a 1 a 2 mod g where g m 1, m 2. Show that if this condition is met, then the solution is unique modulo [m 1, m 2 ]. Proof. If x Z such that x a 1 mod m 1 and x a 2 mod m 2, then n 1, n 2 Z such that x a 1 m 1 n 1 x a 2 m 2 n 2 and hence, a 1 + m 1 n 1 a 2 + m 2 n 2. Conversely, if n 1, n 2 Z such that a 1 + m 1 n 1 a 2 + m 2 n 2, then by letting x be the common value, we see x a 1 mod m 1 and x a 2 mod m 2. Therefore, a simultaneous solution x exists if and only if there exist n 1 and n 2 such that a 1 + m 1 n 1 a 2 + m 2 n 2. This equation is equivalent to m 1 n 1 m 2 n 2 a 2 a 1. By Corollary 1.1.5 in the online notes, such n 1, n 2 exist m 1, m 2 a 2 a 1 a 1 a 2 mod m 1, m 2. To show uniqueness of x modulo [m 1, m 2 ], tae y Z such that x y a 1 mod m 1 and x y a 2 mod m 2. By Theorem 2.3 in the boo, this imlies x y mod [m 1, m 2 ]. Problem. 2.3.47 Let fx be a olynomial with integral coefficients, let Nm denote the number of solutions of the congruence fx mod m, and let φ f m denote the number of integers a, 1 a m, such that fa, m 1. Show that if m, n 1 then φ f mn φ f mφ f n. Show that if > 1 then φ f 1 φ f. Show that φ f N. Conclude that for any ositive integer n, φ f n n N n 1. Show that for an aroriate choice of fx, this reduces to theorem 2.19. Proof. Show that if m, n 1, then φ f mn φ f mφ f n. For any ositive integer, let S {a Z/Z fa, 1} so that φ f #S. Now consider S mn where m, n 1. By the Chinese Remainder theorem we now there is an isomorhism ψ : Z/mnZ Z/mZ Z/nZ. The image of S mn under ψ is ψs mn {ψa Z/mZ Z/nZ fa, mn 1}. Let ψa a 1, a 2. We now that a a 1 mod m and a a 2 mod n, and, by Theorem 2.2 in the boo, fa fa 1 mod m and fa fa 2 mod n. Alying Theorem 2.4 4
we get fa, m fa 1, m and fa, n fa 2, n. Since fa, mn 1 if and only if fa, m fa, n 1, we have fa, mn 1 fa 1, m fa 2, n 1. Hence ψs mn {a 1, a 2 Z/mZ Z/nZ fa 1, m fa 2, n 1} S m S n. We can now restrict the domain of ψ to S mn to establish a bijection between S mn and S m S n. It follows that #S mn #S m #S n and hence φ f mn φ f mφ f n. Next we show that if > 1 then φ f 1 φ f. Notice that fa, 1 fa, 1. Thus φ f number of integers a, 1 a, such that fa, 1. Note that by Theorem 2.2 in the boo: fx y + z fz mod. Theorem 2.4 states: If b c Thus by Theorem 2.4 and Theorem 2.2, we have mod m, then b, m c, m. fx y + z, fz,. Notice that each a such that 1 a can be uniquely written as a a + where 1 a and 1 1. We ve shown fa, 1 fa, 1 fa, 1. So the number of such a is the number of a times the number of all ossible, namely 1, i.e. φ f 1 φ f. Next we show φ f N. φ f number of intergers a 1 a such that fa, 1 fa, 1 fa for some. fa exactly when fa mod. The number of times fa mod is by definition N. Thus φ f N. Finally, let the rime factorization of n be n 1 1 2 φ f n φ f 1 1 φ f 2 2 φ f 2 1 1 1 φ f 1 2 1 2 φ f 2 1 φ f 1 1 1 2 2 2. Since i i, j j 1, 1 1 1 1 N 1 2 1 2 2 N 2 1 N 1 N 1 1 N 2 1 N 1 1 2 2 n n 1 N 1 N 1. 1 1 N 2 2 1 N If we let fx x, then φ f m the number of integers a, 1 a m, such that a, m 1 Thus φ f m φm. N 1 since fx mod x mod which occurs exactly once, when x mod. Thus φm m 1 1, m which is the formula we roved in class. 5