1. Prove that the Mersenne number M13 is a prime; hence, the integer n = 2' (213 – 1) is perfect. 115 implies that the only candidates for prime 01
1. Prove that the Mersenne number M13 is a prime; hence, the integer n = 2' (213 – 1) is perfect. 115 implies that the only candidates for prime 01
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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1
![236
ELEMENTARY NUMBER THEORY
11.
est known even amicable pair whose sum fails to enjoy this feature is 66603025.
and 696630544.
12
PROBLEMS 11.3
= 212 (213 – 1) is
1. Prove that the Mersenne number M13 is a prime; hence, the integer n =1
perfect.
[Hint: Because M13
divisors of M13 are 53 and 79.]
2. Prove that the Mersenne number M19 is a prime; hence, the integer n = 218(219 – 1) is
perfect.
[Hint: By Theorems 11.5 and 11.6, the only prime divisors to test are 191, 457, and 647.]
3. Prove that the Mersenne number M29 is composite.
4. A positive integer n is said to be a deficient number if o (n) < 2n and an abundant number
if o (n) > 2n. Prove each of the following:
(a) There are infinitely many deficient numbers.
[Hint: Consider the integers n = p*, where p is an odd prime and k > 1.]
(b) There are infinitely many even abundant numbers.
[Hint: Consider the integers n = 2* . 3, where k > 1.]
(c) There are infinitely many odd abundant numbers.
[Hint: Consider the integers n = 945 · k, where k is any positive integer not divisible
by 2, 3, 5, or 7. Because 945 = 33 . 5.7, it follows that gcd(945, k)
o(n) = 0 (945)0 (k).]
5. Assuming that n is an even perfect number and d |n, where 1 < d <n, show that d is
deficient.
< 91, Theorem 11.5 implies that the only candidates for prime
%3D
=1 and so
6. Prove that any multiple of a perfect number is abundant.
7. Confirm that the pairs of integers listed below are amicable:
(a) 220 = 2² . 5 · 11 and 284 = 22.71. (Pythagoras, 500 B.C.)
(b) 17296 = 24. 23 47 and 18416 = 24. 1151. (Fermat. 1636)
(c) 9363584 = 27 191 · 383 and 9437056 = 2" ·737^
For a pair of amicable numbers m and n, prove that
rtes, 1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3bf7da54-5138-4b57-a0f9-aec30fa8e8cf%2F530db5e8-304c-47fb-8998-fa2ce0b0844f%2Fss6osil_processed.jpeg&w=3840&q=75)
Transcribed Image Text:236
ELEMENTARY NUMBER THEORY
11.
est known even amicable pair whose sum fails to enjoy this feature is 66603025.
and 696630544.
12
PROBLEMS 11.3
= 212 (213 – 1) is
1. Prove that the Mersenne number M13 is a prime; hence, the integer n =1
perfect.
[Hint: Because M13
divisors of M13 are 53 and 79.]
2. Prove that the Mersenne number M19 is a prime; hence, the integer n = 218(219 – 1) is
perfect.
[Hint: By Theorems 11.5 and 11.6, the only prime divisors to test are 191, 457, and 647.]
3. Prove that the Mersenne number M29 is composite.
4. A positive integer n is said to be a deficient number if o (n) < 2n and an abundant number
if o (n) > 2n. Prove each of the following:
(a) There are infinitely many deficient numbers.
[Hint: Consider the integers n = p*, where p is an odd prime and k > 1.]
(b) There are infinitely many even abundant numbers.
[Hint: Consider the integers n = 2* . 3, where k > 1.]
(c) There are infinitely many odd abundant numbers.
[Hint: Consider the integers n = 945 · k, where k is any positive integer not divisible
by 2, 3, 5, or 7. Because 945 = 33 . 5.7, it follows that gcd(945, k)
o(n) = 0 (945)0 (k).]
5. Assuming that n is an even perfect number and d |n, where 1 < d <n, show that d is
deficient.
< 91, Theorem 11.5 implies that the only candidates for prime
%3D
=1 and so
6. Prove that any multiple of a perfect number is abundant.
7. Confirm that the pairs of integers listed below are amicable:
(a) 220 = 2² . 5 · 11 and 284 = 22.71. (Pythagoras, 500 B.C.)
(b) 17296 = 24. 23 47 and 18416 = 24. 1151. (Fermat. 1636)
(c) 9363584 = 27 191 · 383 and 9437056 = 2" ·737^
For a pair of amicable numbers m and n, prove that
rtes, 1
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