Math 12 Homework 5 Solutions by Luke Gustafson Fall 2003 1. 163 1 2 (mod 2 gives = 2 the smallest rime. 2a. First, consider = 2. We know 2 is not a uadratic residue if and only if 3, 5 (mod 8. By Dirichlet s theorem, there are infinitely many rimes of the form 3 + 8n, so there are infinitely many rimes with = 2 not a uadratic residue. Now consider an odd rime. Let a be a uadratic non-residue modulo. By Dirichlet s theorem, there exists an odd rime r = a + n for some n N. Thus r a (mod, so r is a uadratic non-residue modulo. If r 1 (mod, consider rimes of the form = r + n. If r 3 (mod, then consider = r + 2 + n (notice that r + 2 and must be relatively rime. In each case, it is easy to check 1 (mod. By Dirichlet s theorem, there are infinitely many rimes of this form. Then, use uadratic recirocity to get = ( 1 ( 1( 1 Use the fact that 1 (mod and r (mod : r = (1 = 1 That roves there are infinitely many such that is a uadratic non-residue. 2b. From art (a, we know that for odd rimes 1, the condition that α (mod 1, where α = r if r 1 (mod and α = r + 2 1 if r 3 (mod, is sufficient to ensure 1 is a uadratic non-residue modulo. We may slit this congruence into the euivalent system α (mod and α (mod 1. Also shown in art (a is that α 1 (mod, so our system becomes 1 (mod α (mod 1 1
In the case that 1 = 2, we will use the condition 5 (mod 8 to guarantee that 2 is a uadratic non-residue modulo. In the case that i = 2 for i > 1, we will use the condition 1 (mod 8 to guarantee that 2 is a uadratic residue modulo. For odd rimes i, consider rimes of the form = 1+ i n, where n N. Then we have i = ( 1 ( 1( i 1 Use the fact that 1 (mod and 1 (mod i : 1 = (1 = 1 Hence, it suffices to imose the restriction 1 (mod i to ensure i is a uadratic residue modulo. Euivalently, 1 (mod and 1 (mod i. Now we combine all these conditions on to get a with the desired roerties. We consider the following system of congruences: (1 1 (mod if i 2 for all i (2 5 (mod 8 if 1 = 2 (3 1 (mod 8 if i = 2, i > 1 ( α (mod 1 (5 1 (mod i for i > 1. Notice that conditions (2 and (3 cannot occur simultaneously since we insist all the i are distinct. We show that all rimes satisfying the above system satisfy the conditions of the roblem. By the receding arguments, if i = 2 for some i, then conditions (2 and (3 guarantee that 2 is aroriately a uadratic residue or non-residue. Observe that conditions (1, (2, and (3 guarantee that 1 (mod. Using this fact lus conditions ( and (5, our receding arguments have shown that 1 is a uadratic non-residue and i is a uadratic residue for i > 2, as desired. Now we show that there are infinitely many rimes satisfying our system of congruences. Since the i are distinct, all the moduli are relatively rime. Thus, we aly the Chinese Remainder Theorem to rewrite the system as i x (mod N where N is the roduct of the moduli and x is the uniue solution. Suose, for sake of contradiction, that x and N are not relatively rime. Then there exists a rime dividing x and N. Since divides N, it is eual to 2 or some i. However, our system of congruences guarantees that x is not divisible by 2 (by conditions (1, (2, and (3 or any of the i (by conditions ( and (5. Therefore, x and N are relatively rime. Using Dirichlet s theorem, i 2
there are infinitely many rimes of the form x (mod N, which roves the desired result. 2c. Factor m into rimes a 1 1 a 2 2 ar r. Write a i = 2b i +c i where c i {0, 1}, 1 i r. Then we have m = 2b 1+c 1 1 2b 2+c 2 2 2br+cr r = ( b 1 1 br r 2 c 1 1 cr r Let n = b 1 1 br r. Let 1 2 k = c 1 1 cr r by making the j eual to the i for which c i = 1. The j are distinct because the i are. That gives m in the desired form. 2d. Use art (c to write m in the form n 2 1 k. If m is suare, then it is a uadratic residue for all rimes. If m is not suare, then k 1. By art b, there exists infinitely many ( rimes for which 1 k = 1 2 ( exist infinitely many rimes for which desired. k n 2 1 k = 1. It follows that there = = 1 as n 2 1 k 3a. Let g be a rimitive root modulo. Consider ζ g 1 (mod. Then ζ k g k k (mod, so the order of ζ is the smallest k such that 1 (mod. Letting k < gives an exonent less than 1, which is the smallest ositive exonent for which g is one (since g is a rimitive root. Hence, ζ k 1 (mod when k <. Letting k = gives ζ g 1 1 (mod, so the order of ζ is exactly. g k k 3b. We have ζ 3 1 (mod so (ζ 1(ζ 2 + ζ + 1 0 (mod. Now, ζ 1 (mod because its order is exactly 3. Hence ζ 1 0 (mod, and so ζ 2 + ζ + 1 0 (mod. Therefore, ζ 2 + ζ + 1 3 (mod (2ζ + 1 2 3 (mod.. First, consider the solutions to x 2 1 (mod a for an odd rime. x is a solution if and only if a x 2 1 a (x 1(x + 1. If x 1, then does not divide x + 1, and vice versa. Hence, either a x 1 or a x + 1. Euivalently, x 1 (mod a or x 1 (mod a. That gives us exactly two solutions to the euation x 2 1 (mod a. Now consider the solutions to x 2 1 (mod 2 a. If a = 1, then x 1 (mod 2 is the only solution. If a = 2, one finds the two solutions x ±1 (mod. Next, consider a 3. x is a solution if and only if 2 a (x 1(x + 1. Hence x 1 and x + 1 are consecutive even numbers. That means one of them is divisible by 2 but not. Then, the other number must be divisible by 2 a 1. So, the only ossible solutions are x 1 0 (mod 2 a 1 and x + 1 0 (mod 2 a 1 ; i.e. x ±1 (mod 2 a 1. That gives us four distinct 3
ossibilities modulo 2 a : x ±1 (mod 2 a and x ±1 + 2 a 1 (mod 2 a. Checking, we find that all four of these solutions satisfy x 2 1 (mod 2 a. Hence, there are exactly solutions when a 3. Factor m = a 1 1 a 2 2 a 3 3 ar r. Using the Chinese Remainder Theorem, x 2 1 (mod m if and only if x 2 1 (mod a i i for 1 i r. Moreover, if x is different modulo any i, then x is different modulo m. This establishes an euivalence between the distinct solutions to x 2 1 (mod a i i for each i and the distinct solutions to x 2 1 (mod m. Therefore, the number of solutions to x 2 1 (mod m is r i=1 s i where s i is the number of solutions to x 2 1 (mod a i i. The receding arguments rove that each s i is a ower of two, so it follows that the total number of solutions to x 2 1 (mod m is a ower of two. 5. Define A i to be the ith symmetric sum of the numbers {1, 2,..., 1}; that is, A i = j 1 j 2 j i 1 j 1 <j 2 < <j i 1 The olynomial x 1 1 has solutions 1, 2,..., 1 (mod by Fermat s theorem. Hence, x 1 1 (x 1(x 2 (x +1 (mod. Exanding the right side gives x 1 1 x 1 A 1 x 2 + A 2 x 3 + A 1 (mod (the last term is ositive since 1 is even. Comaring coefficients gives A i 0 (mod for 1 i 2, and A 1 1 (mod. Next, consider the identity (x 1(x 2 (x + 1 = x 1 A 1 x 2 + A 2 x 3 + A 1 Substitute x = to get ( 1! = 1 A 1 2 + A 2 + A 1 Since A 1 = ( 1! it follows that 0 = 1 A 1 2 + A 2 0 = 2 A 1 3 + A 2 We have 2 2 because 5, and i 2 and A i for 1 i 3. Thus, 2 A 1 3 + + A 3 is divisible by 2, which imlies 2 A 2. Now substitute x = 2 into the revious identity. We get (2 1 ( + 1 = (2 1 A 1 (2 2 + 2A 2 + A 1 We have 3 (2 1, 2 (2 i 1 and A i for 1 i 3, and 3 2A 2. Hence, 3 divides all terms on the right side excet A 1. Modulo 3, we get (2 1 ( + 1 A 1 (mod 3
so We have so finally, ( 2 (2 1 ( + 1 ( 1 1 ( 2 = 1 (mod 3 2(2 1 ( + 1 ( 1 1 (2 1 ( + 1 = 2 ( 1 1 (2 1 ( + 1 2 ( 1 1 2 (mod 3 (mod 3 5