hw04-sol-2

CS 70 Discrete Mathematics and Probability Theory Fall 2023 Rao, Tal HW 04 1 Modular Practice Note 6 Solve the following modular arithmetic equations for x and y . (a) 9 x + 5 ≡ 7 ( mod 13 ) . (b) Show that 3 x + 12 ≡ 4 ( mod 21 ) does not have a solution. (c) The system of simultaneous equations 5 x + 4 y ≡ 0 ( mod 7 ) and 2 x + y ≡ 4 ( mod 7 ) . (d) 13 2023 ≡ x ( mod 12 ) . (e) 7 62 ≡ x ( mod 11 ) . Solution: (a) Subtract 5 from both sides to get: 9 x ≡ 2 ( mod 13 ) . Now since gcd ( 9 , 13 ) = 1, 9 has a (unique) inverse mod 13, and since 9 × 3 = 27 ≡ 1 ( mod 13 ) the inverse is 3. So multiply both sides by 9 − 1 ≡ 3 ( mod 13 ) to get: x ≡ 6 ( mod 13 ) . (b) Notice that any number y ≡ 4 ( mod 21 ) can be written as y = 4 + 21 k (for some integer k ). Evaluating y mod 3, we get y ≡ 1 ( mod 3 ) . Since the right side of the equation is 1 ( mod 3 ) , the left side must be as well. However, 3 x + 12 will never be 1 ( mod 3 ) for any value of x . Thus, there is no possible solution. (c) First, subtract the first equation from four times the second equation to get: 4 ( 2 x + y ) − ( 5 x + 4 y ) ≡ 4 ( 4 ) − 0 ( mod 7 ) 8 x + 4 y − 5 x − 4 y ≡ 16 ( mod 7 ) 3 x ≡ 2 ( mod 7 ) Multiplying by 3 − 1 ≡ 5 ( mod 7 ) , we have x ≡ 10 ≡ 3 ( mod 7 ) . Plugging this into the second equation, we have 2 ( 3 )+ y ≡ 4 ( mod 7 ) , so the system has the solution x ≡ 3 ( mod 7 ) , y ≡ 5 ( mod 7 ) . CS 70, Fall 2023, HW 04 1

(d) We use the fact that 13 ≡ 1 ( mod 12 ) . Thus, we can rewrite the equation as x ≡ 13 2023 ≡ 1 2023 ≡ 1 ( mod 12 ) . (e) One way to solve exponentiation problems is to test values until one identifies a pattern. 7 1 ≡ 7 ( mod 11 ) 7 2 ≡ 49 ≡ 5 ( mod 11 ) 7 3 = 7 · 7 2 ≡ 7 · 5 ≡ 2 ( mod 11 ) 7 4 = 7 · 7 3 ≡ 7 · 2 ≡ 3 ( mod 11 ) 7 5 = 7 · 7 4 ≡ 7 · 3 ≡ 10 ≡ − 1 ( mod 11 ) We theoretically could continue this until we the sequence starts repeating. However, notice that if 7 5 ≡ − 1 = ⇒ 7 10 = ( 7 5 ) 2 ≡ ( − 1 ) 2 ≡ 1 ( mod 11 ) . Similarly, 7 60 = ( 7 10 ) 6 ≡ 1 6 ≡ 1 ( mod 11 ) . As a final step, we have 7 62 = 7 2 · 7 60 ≡ 7 2 · 1 = 49 ≡ 5 ( mod 11 ) . 2 Nontrivial Modular Solutions Note 6 (a) What are all the possible perfect cubes modulo 7? In other words, compute the set { x 3 mod 7 | x ∈ Z } . (b) Show that any solution to x 3 + 2 y 3 ≡ 0 ( mod 7 ) must satisfy x ≡ y ≡ 0 ( mod 7 ) . (c) Using part (b), prove that x 3 + 2 y 3 = 7 x 2 y has no non-trivial solutions ( x , y ) in the integers. In other words, there are no integers x and y , that satisfy this equation, except the trivial solution x = y = 0. [ Hint: Consider some nontrivial solution ( x , y ) with the smallest value for | x | (why are we allowed to consider this?). Then arrive at a contradiction by finding another solution ( x ′ , y ′ ) with | x ′ | < | x | .] Solution: (a) Checking by hand, the only perfect cubes modulo 7 are 0, 1, and 6 ≡ − 1: 0 3 ≡ 0 ( mod 7 ) 4 3 ≡ 1 ( mod 7 ) 1 3 ≡ 1 ( mod 7 ) 5 3 ≡ − 1 ( mod 7 ) 2 3 ≡ 1 ( mod 7 ) 6 3 ≡ − 1 ( mod 7 ) 3 3 ≡ − 1 ( mod 7 ) (b) Considering the equation x 3 + 2 y 3 ≡ 0 ( mod 7 ) and considering all cases for x 3 and y 3 , the only way that x 3 + 2 y 3 ≡ 0 ( mod 7 ) is if x 3 ≡ y 3 ≡ 0 ( mod 7 ) . Thus x ≡ y ≡ 0 ( mod 7 ) . CS 70, Fall 2023, HW 04 2

(a) Suppose for the sake of a contradiction that there exists a solution a . First, note that a ≡ 0 ( mod p ) is not a solution. Now, suppose a ̸≡ 0 ( mod p ) . Raise both sides to the p − 1 2 power. Since p ≡ 3 ( mod 4 ) , p − 1 2 is an odd integer. Thus, we have that 1 ≡ a p − 1 ≡ ( a 2 ) p − 1 2 ≡ ( − 1 ) p − 1 2 ≡ − 1 ( mod p ) , since − 1 to an odd integer is − 1. Thus, we have that p | 1 − ( − 1 ) = 2, which is not possible, since p ≡ 3 ( mod 4 ) . Thus, there is no solution to a 2 ≡ − 1 ( mod p ) , as desired. (b) Direction 1 : If p is prime, then the statement holds. For the integers 1 , ··· , p − 1, every number has an inverse. However, it is not possible to pair a number off with its inverse when it is its own inverse. This happens when x 2 ≡ 1 ( mod p ) , or when p | x 2 − 1 = ( x − 1 )( x + 1 ) . Thus, p | x − 1 or p | x + 1, so x ≡ 1 ( mod p ) or x ≡ − 1 ( mod p ) . Thus, the only integers from 1 to p − 1 inclusive whose inverse is the same as itself are 1 and p − 1. We reconsider the product ( p − 1 ) ! = 1 · 2 ··· p − 1. The product consists of 1, p − 1, and pairs of numbers with their inverse, of which there are p − 1 − 2 2 = p − 3 2 . The product of the pairs is 1 (since the product of a number with its inverse is 1), so the product ( p − 1 ) ! ≡ 1 · ( p − 1 ) · 1 ≡ − 1 ( mod p ) , as desired. Direction 2 : The expression holds only if p is prime (contrapositive: if p isn’t prime, then it doesn’t hold). We will prove by contradiction that if some number p is composite, then ( p − 1 ) ! ̸≡ − 1 ( mod p ) . Suppose for contradiction that ( p − 1 ) ! ≡ − 1 ( mod p ) . Note that this means we can write ( p − 1 ) ! as p · k − 1 for some integer k . Since p isn’t prime, it has some prime factor q where 2 ≤ q ≤ n − 2, and we can write p = q · r . Plug this into the expression for ( p − 1 ) ! above, yielding us ( p − 1 ) ! = ( q · r ) k − 1 = q ( rk ) − 1 = ⇒ ( p − 1 ) ! ≡ − 1 ( mod q ) . However, we know q is a term in ( p − 1 ) !, so ( p − 1 ) ! ≡ 0 ( mod q ) . Since 0 ̸≡ − 1 ( mod q ) , we have reached our contradiction. (c) Since p is odd, p − 1 2 is an integer, so p − 1 2 ! exists. Our goal here is to try to connect p − 1 2 ! 2 to ( p − 1 ) !, in order to utilize Wilson’s theorem; if we can show that p − 1 2 ! 2 ≡ ( p − 1 ) ! ( mod p ) , then Wilson’s theorem will directly give us the desired result. Notice that we can express p − 1 2 ! 2 ≡ 1 · 2 ··· p − 3 2 · p − 1 2 · p − 1 2 · p − 3 2 ··· 2 · 1 ( mod p ) CS 70, Fall 2023, HW 04 4

Now, if we can turn the second half of the expansion into the terms p + 1 2 , p + 3 2 , . . . , ( p − 1 ) , then the entire expansion will equal ( p − 1 ) !; to do this, we can negate each term in the second half of the expansion. Further, since p ≡ 1 ( mod 4 ) , we know that p − 1 2 is even, so we’ve introduced an even number of factors of − 1, and these negations will all cancel out. Using this and simplifying, we have ≡ 1 · 2 ··· p − 3 2 · p − 1 2 · − p − 1 2 · − p − 3 2 ··· ( − 2 ) · ( − 1 ) ( mod p ) ≡ 1 · 2 ··· p − 3 2 · p − 1 2 · p − p − 1 2 · p − p − 3 2 ··· ( p − 1 ) ( mod p ) ≡ 1 · 2 ··· p − 3 2 · p − 1 2 · p + 1 2 · p + 3 2 ··· ( p − 1 ) ( mod p ) ≡ ( p − 1 ) ! ( mod p ) ≡ − 1 ( mod p ) where in the last line we applied Wilson’s Theorem. As such, we can conclude that p − 1 2 ! 2 ≡ − 1 ( mod p ) , as desired. 4 Celebrate and Remember Textiles Note 6 Mathematics and computing both owe an immense debt to textiles, where many key ideas origi- nated. Instructions for knitting patterns will tell you to begin by “casting on” the needle some multiple of m plus r , where m is the number of stitches to create one repetition of the pattern and r is the number of stitches needed for the two edges of the piece. For example, in the simple rib stitch pattern below, the repeating pattern is of length m = 4, and you need r = 2 stitches for the edges. + × × ⃝ ⃝ + Pattern of length 4 Edge stitch Edge stitch Thus, to make the final piece wider, you can add as many multiples of the pattern of length 4 as you like; for example, if you want to repeat the pattern 3 times, you need to cast on a total of 3 m + r = 3 ( 4 )+ 2 = 14 stitches (shown below). CS 70, Fall 2023, HW 04 5

and b 3 a 3 ≡ 1 ( mod m 3 ) 28 a 3 ≡ 1 ( mod 5 ) 3 a 3 ≡ 1 ( mod 5 ) → a 3 = 2 . Therefore, x ≡ 6 · 20 · 4 + 2 · 35 · 3 + 2 · 28 · 2 ( mod 140 ) ≡ 102 ( mod 140 ) , so the smallest x that satisfies all three congruences is 102. Therefore we should cast on 102 stitches in order to be able to knit all three patterns into the blanket. 5 Euler’s Totient Theorem Note 6 Note 7 Euler’s Totient Theorem states that, if n and a are coprime, a φ ( n ) ≡ 1 ( mod n ) where φ ( n ) (known as Euler’s Totient Function) is the number of positive integers less than or equal to n which are coprime to n (including 1). (a) Let the numbers less than n which are coprime to n be m 1 , m 2 ,..., m φ ( n ) . Argue that the set { am 1 , am 2 ,..., am φ ( n ) } is a permutation of the set { m 1 , m 2 ,..., m φ ( n ) } . In other words, prove that f : { m 1 , m 2 ,..., m φ ( n ) } → { m 1 , m 2 ,..., m φ ( n ) } is a bijection, where f ( x ) : = ax ( mod n ) . (b) Prove Euler’s Theorem. (Hint: Recall the FLT proof.) Solution: (a) This problem mirrors the proof of Fermat’s Little Theorem, except now we work with the set { m 1 , m 2 , ··· , m φ ( n ) } . Since m i and a are both coprime to n , so is a · m i . Suppose a · m i shared a common factor with n , and WLOG, assume that it is a prime p . Then, either p | a or p | m i . In either case, p is a common factor between n and one of a or m i , contradiction. CS 70, Fall 2023, HW 04 7

We now prove that f is injective. Suppose we have f ( x ) = f ( y ) , so ax ≡ ay ( mod n ) . Since a has a multiplicative inverse ( mod n ) , we see x ≡ y ( mod n ) , thus showing that f is injective. We continue to show that f is surjective. Take any y that is relatively prime to n . Then, we see that f ( a − 1 y ) ≡ y ( mod n ) , so therefore, there is an x such that f ( x ) = y . Furthermore, a − 1 y ( mod n ) is relatively prime to n , since we are multiplying two numbers that are relatively prime to n . (b) Since both sets have the same elements, just in different orders, multiplying them together gives m 1 · m 2 · ... · m φ ( n ) ≡ am 1 · am 2 · ... · am φ ( n ) ( mod n ) and factoring out the a terms, m 1 · m 2 · ... · m φ ( n ) ≡ a φ ( n ) ( m 1 · m 2 · ... · m φ ( n ) ) ( mod n ) . Thus we have a φ ( n ) ≡ 1 ( mod n ) . 6 Sparsity of Primes Note 6 A prime power is a number that can be written as p i for some prime p and some positive integer i . So, 9 = 3 2 is a prime power, and so is 8 = 2 3 . 42 = 2 · 3 · 7 is not a prime power. Prove that for any positive integer k , there exists k consecutive positive integers such that none of them are prime powers. Hint: This is a Chinese Remainder Theorem problem. We want to find n such that ( n + 1 ) , ( n + 2 ) , . . . , and ( n + k ) are all not powers of primes. We can enforce this by saying that n + 1 through n + k each must have two distinct prime divisors. Solution: We want to find n such that n + 1 , n + 2 , n + 3 ,..., n + k are all not powers of primes. We can enforce this by saying that n + 1 through n + k each must have two distinct prime divisors. So, select 2 k primes, p 1 , p 2 ,..., p 2 k , and enforce the constraints n + 1 ≡ 0 ( mod p 1 p 2 ) n + 2 ≡ 0 ( mod p 3 p 4 ) . . . n + i ≡ 0 ( mod p 2 i − 1 p 2 i ) . . . n + k ≡ 0 ( mod p 2 k − 1 p 2 k ) . By Chinese Remainder Theorem, we can calculate the value of n , so this n must exist, and thus, n + 1 through n + k are not prime powers. What’s even more interesting here is that we could select any 2 k primes we want! CS 70, Fall 2023, HW 04 8