5 btat 6(n) = 1, which clearly is impossible by Theorem 7.4.PROBLEMS 7.21 Verify that the equality ø(n) = 4(n + 1) = ¢(n + 2) holds when n =* Show that the integers m == 5186.3k. 638, where k 0, satisfy%3D3k. 568 and n =simultaneously(u)2 = (u)1o (m) = 0 (n), and%3D%3D%3D4. 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) 6(3n) = 3¢(n) if and only if 3| n.(d) ø(3n) = 2¢(n) if and only if 3 X n.(e) 0(n) = n/2 if and only if n =[Hint: Write n =that N = 1.]5. Prove that the equation ø(n) = ¢(n + 2) is satisfied by n = 2(2p – 1) whenever p and2p -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, vn k/2 to obtain ø(n) S-%3D%3D2k for some k 1.U%3D: 2* N, where N is odd, and use the condition ø(n) = n/2 to show%3D%3D%3D22k+1 for k = 1,2, . ]1 ... pr , so 4(n) = 2ko-1 p1-1... pr=1(pı – 1) ...(p, – 1).pr, so o(n) =-1,k1-1|k1/2k, /2P1..(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) 2' · n(1/2).]

Question

5 btat 6(n) = 1, which clearly is impossible by Theorem 7.4.PROBLEMS 7.21 Verify that the equality ø(n) = 4(n + 1) = ¢(n + 2) holds when n =* Show that the integers m == 5186.3k. 638, where k > 0, satisfy%3D3k. 568 and n =simultaneously(u)2 = (u)1o (m) = 0 (n), and%3D%3D%3D4. 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) 6(3n) = 3¢(n) if and only if 3| n.(d) ø(3n) = 2¢(n) if and only if 3 X n.(e) 0(n) = n/2 if and only if n =[Hint: Write n =that N = 1.]5. Prove that the equation ø(n) = ¢(n + 2) is satisfied by n = 2(2p – 1) whenever p and2p -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, vn < p(n) < n.[Hint: Write n = 2ko pNow use the inequalities p – 1 > Jp and k -> k/2 to obtain ø(n) >S-%3D%3D2k for some k > 1.U%3D: 2* N, where N is odd, and use the condition ø(n) = n/2 to show%3D%3D%3D22k+1 for k = 1,2, . ]1 ... pr , so 4(n) = 2ko-1 p1-1... pr=1(pı – 1) ...(p, – 1).pr, so o(n) =-1,k1-1|k1/2k, /2P1..(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 holdsfor%3D12, 14, and 20.(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 kį > 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%3Dk2k1%3D%3D"и2 (и)Ф(и)1 (q)[Hint: Show that t(n)p(n) > 2' · n(1/2).]

Accepted Answer

Euler totient function formula:
p is prime then
ϕ(p)=p-1 
ϕ(pk)=pk-pk-1     ∀k∈ℕ
 
If gcd(a,b)=1…

5. Prove that the equation ø(n) = ¢(n+ 2) is satisfied by n = 2(2p – 1) whenever p and 2p – 1 are both odd primes. %3D -