MATH 310: Homework 2
Due Tuesday, 9/20 in class
Reading: Davenport I.4, I.5
1. Show that xn − y n = (x − y)(xn−1 + xn−2 y + · · · + xy n−2 + y n−1 ).
! n−1
n−1
n−1
X
X
X
n−1−k k
n−k k
(x − y)
x
y
=
x y −
xn−1−k y k+1
k=0
=
k=0
n−1
X
xn−k y k −
k=0
n
X
k=0
xn−` y ` = xn − y n
`=1
2. Show that if 2n − 1 is prime, then n is prime. Is the converse true?
Suppose n is composite, so that n = ab for some a > 1 and b > 1. Then 2ab − 1 =
(2a − 1)((2a )b−1 + · · · + 2a + 1) which is composite because the first factor 2a − 1 > 1,
since a > 1, and the second factor is > 1 since b > 1. Hence, 2n − 1 is composite. Note:
if n = 1 then 2n − 1 = 1. Converse is FALSE because 211 − 1 = 2047 = 23 × 89
3. Show that for n odd, xn + y n = (x + y)(xn−1 − xn−2 y + · · · − xy n−2 + y n−1 ).
! n−1
n−1
n−1
X
X
X
(x + y)
(−1)k xn−1−k y k =
(−1)k xn−k y k +
(−1)k xn−1−k y k+1
k=0
=
k=0
n−1
X
k=0
n
(−1)k xn−k y k −
k=0
n
X
(−1)` xn−` y `
`=1
n n
= x − (−1) y
= xn + y n
since n is odd.
4. Show that if 2n + 1 is prime, then n is a power of 2. Is the converse true?
If n is not a power of 2, then we may write n = ab for some odd b > 1, a > 0. Then
2ab + 1 = (2a + 1)((2a )b−1 − · · · − 2a + 1) which is composite because the first factor
2a + 1 > 1, since a > 0, and the second factor is > 1 since b > 1. Hence, 2n + 1 is
5
composite. Converse is FALSE because 22 + 1 = 4294967297 is divisible by 641.
5. Show that an even perfect number is 1 + 2 + · · · + p for some prime p.
Let n be an even perfect number. By Euler’s result (see BONUS below) n = 2k−1 p for
some prime p = 2k − 1. Hence,
n = 2k−1 p =
(p + 1)p
2k p
=
= 1 + 2 + · · · + p.
2
2
6. (BONUS) Show that every even perfect number is of the form 2k−1 p where p = 2k − 1 is
prime (a.k.a. Mersenne prime).
Let n = 2k−1 pa q b · · · be an even perfect number. In other words, we suppose k ≥ 2,
a > 0, b > 0, . . . where p < q < · · · are the odd prime factors of n, which we have listed
in increasing order for convenience. Then
(2k − 1)(1 + p + · · · + pa )(1 + q + · · · + q b ) = 2k pa q b · · ·
and since 2k − 1 is odd, it divides pa q b · · · , so that dividing out we get
0
0
(1 + p + · · · + pa )(1 + q + · · · + q b ) = 2k pa q b · · ·
for some a0 ≤ a, b0 ≤ b, . . . . At least one of these inequalities is strict, since k ≥ 2.
0
0
Dividing out by pa q b · · · we conclude 2k is a product of factors of the form
b
q + ··· + 1
q b0
each of which is > 1 and at least one of which is > p. Thus, 2k > p.
On the other hand, using
1
1
1
1
1
1 + + ··· + b ··· = 1 + k
1 + + ... a
p
p
q
q
2 −1
and noting the LHS is ≥ 1 + p1 with strict inequality if any of the exponents a, b, . . . are
> 1 or if there are two or more odd prime factors. In the case of strict inequality, we
conclude p1 < 2k1−1 , or equivalently p > 2k − 1, or p ≥ 2k , which is a contradiction.
Thus, there can be at most one odd prime factor and its multiplicity is one.
7. Show that if p and q are odd primes, then pa q b cannot be perfect.
To say that n is perfect means σ(n) = 2n. Suppose pa q b is perfect. Then
(1 + p + · · · + pa )(1 + q + · · · + q b ) = 2pa q b
so that
a b
2p q =
pa+1 − 1
p−1
q b+1 − 1
q−1
<
pa+1
p−1
q b+1
q−1
.
Therefore,
2<
p
p−1
q
q−1
.
x
Since f (x) = x−1
is a decreasing function for all x > 0, its maximum value on [T, ∞) is
T
. If p < q, as we may assume without loss of generality, then p ≥ 3 and q ≥ 5, since
T −1
p
q
3 is the smallest odd prime and 5 is the next smallest. Thus, p−1
≤ /2 and q−1
≤ 5/4 so
that 2 < 15/8, yielding a contradiction.
8. Show that if p is prime and p|ab then p|a or p|b. Did you use the fundamental theorem of
arithmetic (uniqueness)? If so, identify the point(s) in your argument where it is used. If
p - a and p - b, then p does not occur in either prime factorizations of a and b. Multiplying
these prime factorizations gives a factorization of ab into primes which does not involve
p. If p|ab then ab = pq for some q, which can further be factored into primes, so that
ab would have a prime factorization in which one of the prime factors is p and is hence
distinct from the previous one.
The uniqueness part of the fundamental theorem of arithmetic was needed to obtain the
contradiction in the last line.