Download Solutions to Assignment 4 - Real Analysis | MATH 240C and more Assignments Quantitative Techniques in PDF only on Docsity!
Math 240 Solutions to Assignment 4 [email protected]
This assignment carries a total of twenty points.
Question 1. [6]
(a) Find the gcd of 35 and 343. (b) Find a formula in k for the lcm of 2k + 1 and 2k − 1. (c) Find the gcd of k + 1 and k^2 + 1 when k is a positive odd integer.
Solution 1.
A mark is deducted for each question where you do not justify your answer.
(a) The Euclidean Algorithm is applied:
gcd(35, 343) = gcd(35, 28) = gcd(7, 28) = gcd(7, 0) = 7.
(b) We use the formula lcm(a, b) = ab/gcd(a, b). By the Euclidean Algorithm,
gcd(2k − 1 , 2 k + 1) = gcd(2k − 1 , 2) = gcd(1, 2) = gcd(1, 0) = 1.
Therefore lcm(2k − 1 , 2 k + 1) = 4k^2 − 1. Alternatively, observe that if d is a divisor of 2 k − 1 and 2k + 1, then d|(2k − 1) + 2 so d|2 which means d = 1 or d = 2. But d 6 = 2 since 2k − 1 and 2k + 1 are odd. So d = 1, which means gcd(2k − 1 , 2 k + 1) = 1. (c) If we do long division (or by inspection), we obtain k^2 + 1 = (k − 1)(k + 1) + 2 so the remainder when k^2 + 1 is divided by k + 1 is 2. The Euclidean Algorithm gives
gcd(k + 1, k^2 + 1) = gcd(k + 1, 2) = 2
since k + 1 is even.
Question 2. Prove that for any positive integers x, y and any prime number p,
(x + y)p^ ≡ xp^ + yp^ mod p
Find an example of integers x, y, n such that (x + y)n^6 ≡ xn^ + yn^ mod n. [3]
Solution 2.
By Fermat’s Theorem, p|(x + y)p^ − (x + y); equivalently, saying (x + y)p^ ≡ (x + y) mod p. Therefore the left hand side is x + y mod p whereas the right hand side is x mod p + y mod p which is x + y mod p. Alternatively, use the binomial theorem
(x + y)p^ =
∑^ p
k=
p k
xkyp−k
and from a lemma in class we know
(p k
≡ 0 mod p when 1 ≤ k ≤ p − 1. So mod p the left hand side is, as required, (^) ( p 0
xp^ +
p p
yp^ = xp^ + yp
There are many examples of x, y, n for which (x + y)n^ ≡ xn^ + yn^ mod n does not hold if n is not prime. For example, if n = 4 and x = 1 and y = 3, then
(x + y)n^ = 4^4 ≡ 0 mod 4
whereas x^4 + y^4 ≡ 14 + 3^4 ≡ 14 + (−1)^4 ≡ 2 mod 4
and so (x + y)^4 6 ≡ x^4 + y^4.
Question 3. Give a proof of or counterexample to each statement, where p is a prime and
x, y are positive integers: [6]
(a) If x−^1 ≡ y−^1 mod p then x ≡ y mod p. (b) If x^2 ≡ y^2 mod p then x ≡ y mod p. (c) For any integer x, xp^ ≡ x mod p.
Solution 3.
(a) True. Multiply both sides by xy to get y ≡ x mod p, as required. (b) False. This is since x ≡ −y mod p is possible: For example, take x = 1 and y = p − 1: then 1^2 ≡ (p − 1)^2 mod p but if p > 2 then 1 6 ≡ p − 1 mod p. (c) True. This is Fermat’s Theorem.
Question 4. Solve each of the modular equations for x: [6]
(a) 7 x + 11 ≡ 2 mod 13. (b) − 2 x − 3 ≡ 4 mod 9. (c) x^2 ≡ −1 mod 17.
Solution 4.
(a) This is the same as 7x ≡ −9 mod 13 i.e. 7x ≡ 4 mod 13. So x ≡ 7 −^1 · 4 mod 13; we know that 7−^1 exists since 13 is prime. To find 7−^1 , we use the Euclidean Algorithm:
gcd(7, 13) = gcd(7, 13 − 7) = gcd(7 − (13 − 7), 13 − 7) = gcd(1, 6).
Therefore 7 − (13 − 7) = 1 which implies 2 · 7 − 13 = 1 and 7−^1 ≡ 2 mod 13. The answer is therefore x = 8. You are required to show the Euclidean Algorithm here, otherwise a mark is deducted. (b) − 2 ≡ 7 mod 9 so the equation is 7x ≡ 7 mod 9. Since gcd(7, 9) = 1, 7−^1 exists. So x = 7−^1 · 7 = 1 (we don’t even need to work out 7−^1 ). (c) This is the same as x^2 − 16 ≡ 0 mod 17. Now 17 is prime, so (x − 4)(x + 4) ≡ 0 mod 17 implies x − 4 ≡ 0 mod 17 or x + 4 ≡ 0 mod 17 (it is important to note that 17 is prime, otherwise we can’t say this). Therefore x ≡ 4 or x ≡ −4 mod 17. So x = 4 or x = −4 is the answer.
