Download MATH 406: Homework 7.3 Solutions 1. Find the five smallest

MATH 406: Homework 7.3 Solutions
1. Find the five smallest positive deficient numbers and justify why each is deficient.
Solution: We have
n = 1 σ(1) = 1 < 2(1)
n = 2 σ(2) = 3 < 2(2)
n = 3 σ(3) = 4 < 2(3)
n = 4 σ(4) = 7 < 2(4)
n = 5 σ(5) = 6 < 2(5)
2. Find the five smallest positive perfect numbers and justify why each is perfect.
Solution: We have
n=6
σ(6) = σ(2 · 3) = 12 = 2(6)
n = 28
σ(28) = σ(22 · 7) = 56 = 2(28)
n = 496
σ(496) = σ(24 · 31) = 992 = 2(496)
n = 8128
σ(8128) = σ(26 · 127) = 16256 = 2(8128)
n = 33550336 σ(33550336) = σ(212 · 8191) = 67100672 = 2(33550336)
3. Find the five smallest positive abundant numbers and justify why each is abundant.
Solution: We have
n = 12 σ(12) = 28 > 2(8)
n = 18 σ(18) = 39 > 2(18)
n = 20 σ(20) = 42 > 2(20)
n = 24 σ(24) = 60 > 2(24)
n = 30 σ(30) = 72 > 2(30)
4. Show that if a and b are positive integers and p and q are distinct odd primes then n = pa q b is deficient.
Solution: Assume p > q > 2. We must show that σ(pa q b ) < 2pa q b . Observe that
b+1
a+1 b+1 a+1
q
−1
p
q
p
q
p
−1
a b
<
=
pa q b
σ(p q ) =
p−1
q−1
p−1
q−1
p−1
q−1
q
p
so it suffices to show that p−1
q−1 < 2. Rewriting this desired inequality as
p
p−1
q
q−1
<2
pq < 2(p − 1)(q − 1)
pq < 2pq − 2p − 2q + 2
0 < pq − 2p − 2q + 2
2 < (p − 2)(q − 2)
Since p > q > 2 we have (p − 2)(q − 2) > (5 − 2)(3 − 2) = 3 > 2 so since the final inequality is true and
the chain of inequalities is reversible the first must be true. So finally we can say
p
q
a b
σ(p q ) <
pa q b < 2pa q b
p−1
q−1
and so pa q b is deficient.
5. Prove that a perfect square cannot be a perfect number.
Solution: Since n is a perfect square we have n = 2k pk11 ...pki i where the pi are the primes other than
2 and all the k are even.
First we’ll count the number of odd divisors. Since these are the ones with no 2 in them we have
(k1 + 1)...(ki + 1) of them. Since each ki is even this product is odd. Therefore n has an odd number
of odd divisors.
But now observe that the sum of all the divisors of n is the sum of the odd divisors plus the sum of
the even divisors. Since we have an odd number of odd divisors they sum to an odd number. The even
divisors sum to an even number. Thus the sum of all the divisors is odd and cannot equal 2n. Thus n
is not perfect.
6. Prove that if n is not a perfect square then τ (n) is even.
Solution: We’ll prove that τ (n) odd implies n is a perfect square. Suppose n = pk11 ...pki i then
τ (n) = (k1 + 1)...(ki + 1). For this to be odd we need all the kj + 1 odd meaning all the ki even. Thus
2
ki = 2ji for all i and then n = pj11 ...pji i .
7. Prove that there are infinitely many even abundant numbers.
Solution: Let n = 22 32 p where p is any prime not equal to 2 or 3. Then
σ(n) = σ(22 32 p)
= σ(22 )σ(32 )σ(p)
= (1 + 2 + 22 )(1 + 3 + 32 )(p + 1)
= (7)(13)(p + 1)
7
13
> 22
32
p
4
9
91
>
n
36
> 2n
so that n is abundant. Since there are infinitely many choices for p there are infinitely many abundant
numbers.
8. Use Theorem 7.12 to determine whether each of the following Mersenne numbers is a Mersenne prime:
(a) M7
7
Solution: The only divisors greater
√ than 1 of M7 = 2 − 1 = 127 must have the form 2(7)k + 1,
that is, 14k + 1 for k ≥ 1. Since 127 ≈ 11.27 and we only need to try numbers smaller than
this, there are no such numbers, so M7 is prime.
(b) M11
11
Solution: The only divisors greater
√ than 1 of M11 = 2 −1 = 2047 must have the form 2(11)k+1,
that is, 22k + 1 for k ≥ 1. Since 2047 ≈ 45.24 and we only need to try numbers smaller than
this, we try 23 and 45. We find 2047 = (23)(89) so M11 is composite.
(c) M17
Solution: The only divisors greater than 1√of M17 = 217 − 1 = 131071 must have the form
2(17)k + 1, that is, 34k + 1 for k ≥ 1. Since 131071 ≈ 362.04 and we only need to try numbers
smaller than this, we try 35, 69, 103, and 137. None of these work and so M17 is prime.
(d) M29
Solution: The only divisors greater than 1 of√M29 = 229 − 1 = 536870911 must have the form
2(29)k + 1, that is, 58k + 1 for k ≥ 1. Since 536870911 ≈ 23170.47 and we only need to try
numbers smaller than this, we try 59, 117, 175, 233 and luckily we get to stop there because 233
is a factor and so M29 is composite.