8. Prove that if the integer n has r distinct odd prime factors, then 2" | ¢(n).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
8
![b
r (n) = 1, which clearly is impossible by Theorem 7.4.
PROBLEMS 7.2
1 Calculate (1001), ø(5040), and ø(36,000).
1 Verify that the equality ø(n) = 4(n + 1) = ¢(n + 2) holds when n =
* Show that the integers m = 3* . 568 and n = 3k . 638, where k > 0, satisfy
= 5186.
%3D
simultaneously
(u)2 = (u)1
o (m) = 0 (n), and
%3D
%3D
%3D
A. Establish each of the assertions below:
(a) If n is an odd integer, then o(2n) = 4(n).
h) If n is an even integer, then ø(2n) = 2(n).
(c) ¢(3n) = 3¢(n) if and only if 3 | n.
(d) ø(3n) = 2¢(n) if and only if 3 X n.
(e) o(n) = n/2 if and only if n =
[Hint: Write n =
that N = 1.]
S-
%3D
%3D
2k for some k > 1.
%3D
: 2* N, where N is odd, and use the condition ø(n) = n/2 to show
%3D
= 1.]
5. Prove that the equation o(n) = ¢(n + 2) is satisfied by n = 2(2p – 1) whenever p and
2p - 1 are both odd primes.
6. Show that there are infinitely many integers n for which (n) is a perfect square.
[Hint: Consider the integers n =
7. Verify the following:
(a) For any positive integer n, n < ¢(n) < n.
[Hint: Write n = 2ko p
Now use the inequalities p – 1 > Jp and k -> k/2 to obtain ø(n) >
2ko-1 pi'
%3D
22k+1 for k = 1,2, . ]
%3D
1 ... pr , so 4(n) = 2ko-1 p1-1... pr=1(pı – 1) ..(p, – 1).
p-(pı – 1)..(p, – 1).
pr, so o(n) =
-1,k1-1
-1„k1/2
..
(b) If the integer n > 1 has r distinct prime factors, then ø(n) > n/2".
(c) If n > 1 is a composite number, then ø(n) <n – /n.
[Hint: Let p be the smallest prime divisor of n, so that p < n. Then
$(n) < n(1 – 1/p).]
8. Prove that if the integer n has r distinct odd prime factors, then 2" | 4(n).
9. Prove the following:
(a) If n and n + 2 are a pair of twin primes, then ø(n + 2) = (n) + 2; this also holds
for
-
%3D
12, 14, and 20.
= u JoJ
(b) If p and 2p + 1 are both odd primes, thenn = 4p satisfies $(n + 2) = ¢(n)+2.
10. If every prime that divides n also divides m, establish that ø(nm)= nø(m); in particular,
$(n²) = nø(n) for every positive integer n.
11. (a) If ø(n)|n – 1, prove that n is a square-free integer.
[Hint: Assume that n has the prime factorization n = p p... pr, where ki > 2.
Then p1|$(n), whence p1 |n - 1, which leads to a contradiction.]
(b) Show that if n = 2k or 2k3i, with k and j positive integers, then ø(n)|n.
12. If n = p p... pk, derive the following inequalities:
(a) o (n)p(n) > n²(1 – 1/p})(1 – 1/pź).. (1 – 1/p²).
%3D
%3D
%3D
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k2
k1
%3D
%3D
"и2 (и)Ф(и)1 ()
[Hint: Show that t(n)#(n) > 2" · n(1/2).]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fadf70025-15e9-4d93-968e-9d86049b61bd%2F3cd1bb4c-587e-4327-b92c-880acb7273ee%2F3yfg6ie.jpeg&w=3840&q=75)
Transcribed Image Text:b
r (n) = 1, which clearly is impossible by Theorem 7.4.
PROBLEMS 7.2
1 Calculate (1001), ø(5040), and ø(36,000).
1 Verify that the equality ø(n) = 4(n + 1) = ¢(n + 2) holds when n =
* Show that the integers m = 3* . 568 and n = 3k . 638, where k > 0, satisfy
= 5186.
%3D
simultaneously
(u)2 = (u)1
o (m) = 0 (n), and
%3D
%3D
%3D
A. Establish each of the assertions below:
(a) If n is an odd integer, then o(2n) = 4(n).
h) If n is an even integer, then ø(2n) = 2(n).
(c) ¢(3n) = 3¢(n) if and only if 3 | n.
(d) ø(3n) = 2¢(n) if and only if 3 X n.
(e) o(n) = n/2 if and only if n =
[Hint: Write n =
that N = 1.]
S-
%3D
%3D
2k for some k > 1.
%3D
: 2* N, where N is odd, and use the condition ø(n) = n/2 to show
%3D
= 1.]
5. Prove that the equation o(n) = ¢(n + 2) is satisfied by n = 2(2p – 1) whenever p and
2p - 1 are both odd primes.
6. Show that there are infinitely many integers n for which (n) is a perfect square.
[Hint: Consider the integers n =
7. Verify the following:
(a) For any positive integer n, n < ¢(n) < n.
[Hint: Write n = 2ko p
Now use the inequalities p – 1 > Jp and k -> k/2 to obtain ø(n) >
2ko-1 pi'
%3D
22k+1 for k = 1,2, . ]
%3D
1 ... pr , so 4(n) = 2ko-1 p1-1... pr=1(pı – 1) ..(p, – 1).
p-(pı – 1)..(p, – 1).
pr, so o(n) =
-1,k1-1
-1„k1/2
..
(b) If the integer n > 1 has r distinct prime factors, then ø(n) > n/2".
(c) If n > 1 is a composite number, then ø(n) <n – /n.
[Hint: Let p be the smallest prime divisor of n, so that p < n. Then
$(n) < n(1 – 1/p).]
8. Prove that if the integer n has r distinct odd prime factors, then 2" | 4(n).
9. Prove the following:
(a) If n and n + 2 are a pair of twin primes, then ø(n + 2) = (n) + 2; this also holds
for
-
%3D
12, 14, and 20.
= u JoJ
(b) If p and 2p + 1 are both odd primes, thenn = 4p satisfies $(n + 2) = ¢(n)+2.
10. If every prime that divides n also divides m, establish that ø(nm)= nø(m); in particular,
$(n²) = nø(n) for every positive integer n.
11. (a) If ø(n)|n – 1, prove that n is a square-free integer.
[Hint: Assume that n has the prime factorization n = p p... pr, where ki > 2.
Then p1|$(n), whence p1 |n - 1, which leads to a contradiction.]
(b) Show that if n = 2k or 2k3i, with k and j positive integers, then ø(n)|n.
12. If n = p p... pk, derive the following inequalities:
(a) o (n)p(n) > n²(1 – 1/p})(1 – 1/pź).. (1 – 1/p²).
%3D
%3D
%3D
%3D
k2
k1
%3D
%3D
"и2 (и)Ф(и)1 ()
[Hint: Show that t(n)#(n) > 2" · n(1/2).]
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