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Mathematics 2124b – 2014
Some Solutions 7.1
7.1.11 Note that (n4 + 3n2 + 1) − n(n3 + 2n) = n2 + 1, then note that (n3 + 2n) − n(n2 +
1) = n, and finally that (n2 + 1) − n(n) = 1. It follows that any common divisor of
n4 + 3n2 + 1 and n3 + 2n is also a divisor of n2 + 1, and of n, and finally of 1. Therefore
gcd(n3 + 2n, n4 + 3n2 + 1) = 1 for all integers n, as required.
This proof can be given more formally by applying the Euclidean algorithm to the two
polynomials to find the gcd.
7.1.13 (a) We substitute: 17(2 + 11t) + 11(−3 − 71t) = 34 + 17 · 11t − 33 − 11 · 17t = 1, as
required.
(b) We know that x = 2, y = −3 is a solution. Now suppose that x = a and y = b
give some other solution, i.e. 17a + 11b = 1 where a, b ∈ Z. We must show that there
is some t ∈ Z with a = 2 + 11t, b = −3 − 17t. We have
17(2) + 11(−3) = 1
17a + 11b = 1
It follows that 17(a − 2) + 11(b + 3) = 0, or 17(a − 2) = −11(b + 3). Therefore 11
divides 17(a − 2). Since 17 ⊥ 11, we must have 11 | (a − 2); thus there is some t ∈ Z
with a − 2 = 11t, or a = 2 + 11t. Substituting for a, using this value of t, we also find
17(2 + 11t − 2) = −11(b + 3), and it follows that b = −3 − 17t as required.
7.1.15 We give a proof by induction. For n = 1, the two consecutive Fibonacci numbers are
fn = f1 = 1 and fn+1 = f2 = 1; we have gcd(1, 1) = 1 as required.
For the induction step, we assume that gcd(fn , fn+1 ) = 1, where n ≥ 1; we have
to prove gcd(fn+1 , fn+2 ) = 1. From the Fibonacci recurrence, fn+2 = fn+1 + fn , or
fn+2 − fn+1 = fn . Therefore any common divisor of fn+2 and fn+1 is also a divisor of
fn (and fn+1 ). By induction hypothesis, the only positive common divisor of fn and
fn+1 is 1; therefore the only positive common divisor of fn+2 and fn+1 is 1 also and we
are done.
7.1.18 Let s = p + q where p and q are consecutive primes, i.e. p and q are prime, p < q, and
there is no prime r with p < r < q. If s = 2r, where r ∈ N, then r = s/2 = (p + q)/2,
i.e. r is the average of p and q, so p < r < q. Therefore r is not prime.
7.1.19 By experimentation you can find long sequences of consecutive composite numbers,
for example 122, 123, 124, 125, 126. In general, note that n! + i is divisible by i for
2 ≤ i ≤ n, so each such number has a proper divisor and is therefore composite. That
is, the sequence n! + 2, n! + 3, n! + 4, . . . , n! + n is a sequence of consecutive composite
numbers of length n − 1. It follows that we can find arbitrarily long sequences of
consecutive composite numbers.
1
7.1.20 We try the first few case, hoping for inspiration (of course, it’s understood that n ≥ 2,
otherwise the problem doesn’t make sense):
1+
1
2
1 1
1+ +
2 3
1 1 1
1+ + +
2 3 4
1 1 1 1
1+ + + +
2 3 4 5
1+
+
+
+
+
1
2
1
3
1
4
1
5
1
6
=
=
=
=
=
3
,
2
3 1
11
+ = ,
2 3
6
11 1
25
+ = ,
6
4
12
25 1
137
+ =
,
12 5
60
137 1
147
+ =
.
60
6
60
So far, we have found no integer values for the sum, so we take heart that the statement
of the problem may be correct.
There is one potentially useful pattern present: in lowest terms the sum is always (at
odd
. If we can show that this pattern persists, then the sum
least, so far) of the form
even
can never be an integer.
Let k be the largest exponent for which 2k ≤ n; in other words, k = blog2 nc. Then in
the sequence 1, 2, . . . , n there is exactly one which is divisible by 2k , namely 2k itself
(the next integer after 2k which is divisible by 2k would be 2k+1 , which is larger than
n, by choice of k; and certainly no positive integer smaller than 2k is divisible by 2k ).
The least common multiple of all the denominators (the “lowest common denomina1
1 1 1
tor”) in the sum 1 + + + + · · · + will be 2k o where o is odd; In particular, note
2 3 4
n
that this lcm will be even.
1
1
even
All of the fractions , except for k , will be of form k when written in terms of this
j
2
2 o
common denominator. When all the numerators are added, all but one will be even;
therefore the sum is odd. Any cancellations required to put the sum in lowest terms
will preserve this property, since 2 is not a factor of the numerator.
!
p
p!
7.1.21
=
. For 1 ≤ k ≤ p − 1, none of the factors in the denominator is p, so
k!(p − k)!
k
after the cancellation takes place, there will be a factor p remaining.
7.1.22 Each factor of 10 in 1 · 2 · 3 · · · n will produce a zero in n!. Since 10 = 2 · 5, we need
to count the number of twos and fives that occur in 1 · 2 · 3 · · · n. There are more twos
than fives, so in fact it suffices to count the fives. That is, we need to count how many
fives occur as factors of 1, 2, · · · , n. We have to be careful since some numbers, such
as 25, have more than one five. To begin with, we count those numbers that have at
least one factor of 5; that is, we count the set {5, 10, 15, 20, 25,$. . .}
% where we stop at
n
the last multiple of 5 less than or equal to n. This count gives
.
5
2
We have not yet counted all the fives that appear in those of 1, 2, . . . , n which are
multiples of
$ 25%— in fact, we have counted only one of the fives in each of these cases.
n
counts the second five that appears in each multiple of 25.
Therefore
25
$
Of course, some numbers may have even more fives as factors;
$
third 5 that occurs in multiples of 125;
n
625
n
125
%
will count the
%
will count the fourth 5 that occurs in
multiples of 625, and so on.
In general, the total number of fives occurring as factors of 1 · 2 · 3 · · · n is
$ %
$
%
$
%
$
%
n
n
n
n
+
+
+
+ ···
5
25
125
625
where we don’t have to specify an upper limit for the sum, since eventually (for 5i > n)
the terms will all be zero.
In particular, for n = 1000, we have
$
%
$
%
$
%
$
%
1000
1000
1000
1000
+
+
+
= 200 + 40 + 8 + 1 = 249
5
25
125
625
as required.
7.1.23 This one is based on the same ideas as 7.1.22: first recall that
pr
k
!
=
(pr )!
.
k!(pr − k)!
The number of factors p in the numerator is
pr ! pr ! pr !
+ 2 + 3 + ···.
p
p
p
The number of factors of p in the denominator is
$
%
$
%
$
%
k!
k!
k!
+ 2 + 3 + ··· +
p
p
p
$
(pr − k)!
(pr − k)!
(pr − k)!
+
+
+ ···.
p
p2
p3
%
$
%
$ %
$
$ %
%
$
%
x
y
x+y
We next turn to a useful property of the floor function:
+
≤
for all
z
z $
z
$ %
$
%
%
k!
(pr − k)!
k! + (pr − k)!
x, y, z ∈ R. (Proof: exercise.) In particular
+
≤
.
pj
pj $
pj$
$ %
%
%
k!
(pr − k)!
(pr )!
We also have k! + (pr − k)! ≤ (pr )! (exercise); thus
+
≤
.
pj
pj
pj
Furthermore, for at least one value of j (namely j = r) we have strict inequality.
pr
It follows that the number of times that p occurs in the numerator of
is greater
k
than! the number of times p occurs in the denominator, and therefore p is a divisor of
pr
.
k
!
3
7.1.24 Suppose n | fm for some integers n > 0 and m ≥ 0. Consider fm+1 , fm+2 , fm+3 , . . ..
We have fm+2 = fm+1 + fm ; then fm+3 = fm+2 + fm+1 = 2fm+1 + fm ; then fm+4 =
fm+3 + fm+2 = 3fm+1 + 2fm ; then fm+5 = fm+4 + fm+3 = 5fm+1 + 3fm ; then fm+6 =
fm+5 + fm+4 = 8fm+1 + 5fm , and so on. The coefficients that appear are themselves
Fibonacci numbers: fm+k = fk fm+1 + fk−1 fm (proof: induction exercise). It follows
that fm+m = fm fm+1 + fm−1 fm , so if n | fm then n | f2m also; and then there are
infinitely many Fibonacci multiples of n.
7.1.25 Let 3 = p1 < p2 < p3 < . . . < pr be primes each of form p = 4k + 3; our task
is to show that there must be some prime of form 4k + 3 which is not in the list
p1 , p2 , . . . , pr . We may assume that r > 1 since we know that 3 and 7 are primes.
Consider N = 4p2 p3 · · · pr + 3; since N is odd, all prime factors of N are of form 4k + 1
or 4k + 3 (i.e. all prime factors of N must be congruent to 1 or 3 modulo 4). None
of p1 , p2 , p3 , . . . , pr is a factor of N , since when n is divided by any of p2 , p3 , . . . , pr ,
a remainder of 3 results; in addition note that if 3 | N then 3 | 4p2 p3 · · · pr , which is
impossible since 3 ⊥ 4, 3 ⊥ p2 , . . . , 3 ⊥ pr .
The product of numbers congruent to 1 modulo 4 is again congruent to 1 modulo 4.
However, N ≡ 3 (mod 4). Therefore not all prime factors of N are of form 4k + 1, so
there must be at least one prime factor of N of form 4k + 3. As noted above, this will
not be in the list p1 , p2 , . . . , pr ; we conclude that there is some prime of form 4k + 3
not on this list, as required.
7.1.26 Similar to 7.1.25. Note that numbers of form 6k, 6k + 2, 6k + 3, 6k + 4 cannot be prime
for k ≥ 1. Suppose a list of primes of form 6k + 5 is given. What can you say about
the product of numbers each of which is of form 6k + 1?
7.1.28 (a) This is true; that is, the product of two consecutive positive integers cannot be
a perfect square. Proof: consecutive numbers are relatively prime, but uv = n2 with
u ⊥ v implies that both u and v are squares. This does not happen for u, v consecutive
positive integers since the difference between consecutive squares m2 and (m + 1)2 is
2m + 1 > 1 for m ≥ 1.
(b) This is also true, i.e. the product of 3 consecutive positive integers cannot be a
perfect square. Let the three integers be a, a + 1, a + 2. Since any two consecutive
integers are relatively prime, we have a ⊥ (a+1) and (a+1) ⊥ (a+2). If a(a+1)(a+2)
is a square it follows that a + 1 is a perfect square, b2 , so we may write the product
as (b2 − 1)(b2 )(b2 + 1). Then (b2 − 1)(b2 + 1) = b4 − 1 must also be a perfect square.
This is impossible since b4 itself is a perfect square, and we have seen in (a) that two
consecutive positive integers cannot both be perfect squares.
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