Download Raji, Exercises 4.1: 1. Determine whether the arithmetic functions f

Raji, Exercises 4.1:
1. Determine whether the arithmetic functions f (n) = n! and g (n) = n=2
are completely multiplicative or not.
For f , note that f (2) = 2 and f (3) = 6 and f (2 3) = 720 6= 2 6, so f
is not even multiplicative.
For g, not that g (1 1) = 1=2 6= 1=4 = g (1) g (1), so g is not multiplicative
either.
2. De…ne the arithmetic function g (n) by g (1) = 1 and g (n) = 0 if n > 1.
Prove that g is multiplicative.
In fact, g is completely multiplicative. If ab = 1, then f (ab) = f (a) f (b) =
1, and if ab > 1, then f (ab) = f (a) f (b) = 0.
Did you catch the misprint in Example 33?
Raji, Exercises 4.2:
1. Find ' (256) and ' (2 3 5 7 11).
As 256 = 28 , it follows that ' (256) = 28
27 = 27 (2
1) = 27 = 128.
Curious how Raji writes multiplication as a dot on the line rather than a
centered dot. Because ' is a multiplicative function,
' (2 3 5 7 11) = ' (2) ' (3) ' (5) ' (7) ' (11) = 1 2 4 6 10 = 480
2. Show that ' (5186) = ' (5187).
I guess we have to factor those numbers into primes: 5186 = 2 2593 and
5187 = 3 7 13 19. So ' (5186) = 1 2592, and ' (5187) = 2 6 12 18 = 2592.
3. Find all positive integers n such that ' (n) = 6.
How far do we have to look? If n is divisible by a prime p, then ' (n) is
divisible by p 1, so n can’t be divisible by 5, or by any prime greater
than 7. If n is divisible by 23 , then ' (n) is divisible by 4. If n is divisible
by 33 , then ' (n) is divisible by 9. If n is divisible by 72 , then ' (n) is
divisible by 7. So the only possibilities for n are 2a 3b 7c where 0 a 2,
0 b 2, and 0 c 1. These are the 18 numbers
1; 2; 4; 3; 6; 12; 9; 18; 36; 7; 14; 28; 21; 42; 84; 63; 126; 252:
The values of ' on these numbers are
1; 1; 2; 2; 2; 4; 6; 6; 24; 6; 6; 12; 12; 12; 24; 36; 36; 72:
So the positive integers n such that ' (n) = 6 are 9, 18, 7, and 14.
4. Show that if n is a positive integer, then ' (2n) = ' (n) if n is odd.
If n is odd, then gcd (2; n) = 1, so ' (2n) = ' (2) ' (n) = 1 ' (n).
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5. Show that if n is a positive integer, then ' (2n) = 2' (n) if n is even.
If n is even, then n = 2m k where m > 0 and k is odd. So
' (2n)
=
' 2m+1 k = ' 2m+1 ' (k) = 2m ' (k)
=
2 2m
1
' (k) = 2' (2m ) ' (k) = 2' (n)
6. Show that if n is an odd integer, then ' (4n) = 2' (n).
I suppose that means “n is an odd positive integer”. Isn’t this easy? If n
is odd, then ' (4n) = ' (4) ' (n) = 2' (n).
7. Find the sum of positive integer divisors and the number of positive integer
divisors of 35.
So we are supposed to calculate (35) and (35). The positive integer
divisors of 35 are 1, 5, 7, and 35. So (35) = 48, and (35) = 4.
8. Find the sum of positive integer divisors and the number of positive integer
divisors of 23 34 53 73 13.
Use Theorem 43 to see that the sum is
24
2
1 35
1 3
1 54
1 5
1 74
1 7
1 132
1 13
1
1
The …ve factors are 15, 121, 156, 400, 14, and 15 121 156 400 14 =
1585 584 000
The number of divisors is given by Theorem 45. It is
4 5 4 4 2 = 640
9. Which positive integers have an odd number of positive divisors?
The squares. This follows from Theorem 45. For the number of divisors of
n to be odd, each of the exponents of the primes in the prime factorization
of n has to be even, which says that n is square.
Alternatively, if n is not a square, then the divisors of n come in pairs: d
and n=d, which must be distinct because n = d n=d is not a square. But
if n = a2 , only the divisor a is not paired with a di¤erent divisor, so there
are an odd number of divisors.
10. Which positive integers have exactly two positive divisors?
Must be the primes. That could be taken as a de…nition of a prime. Every
positive integer n is divisible by 1 and n. If 1 = n, then we don’t call n
a prime. If there is a third divisor of n, then we don’t call n a prime.
Otherwise, we call n a prime.
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