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2.2
Perfect Cubes
Problem 2.2.5.
Find all positive perfect cubes that are not divisible by
10
such
that the number obtained by erasing the last three digits is also a perfect cube.
Solution.
We have
(
10
m
+
n
)
3
=
1000
a
3
+
b
, where 1
≤
n
≤
9 and
b
<
1000.
The equality gives
(
10
m
+
n
)
3
−
(
10
a
)
3
=
b
<
1000
,
so
(
10
m
+
n
−
10
a
)
[
(
10
m
+
n
)
2
+
(
10
m
+
n
)
·
10
a
+
100
a
2
]
<
1000
.
Since
(
10
m
+
n
)
2
+
(
10
m
+
n
)
·
10
a
+
100
a
2
>
100, we obtain 10
m
+
n
−
10
a
<
10; hence
m
=
a
.
If
m
≥
2, then
n
(
300
m
2
+
30
mn
+
n
2
) >
1000, false.
Then
m
=
1 and
n
(
300
+
30
n
+
n
2
) <
1000; hence
n
≤
2. For
n
=
2, we
obtain 12
3
=
1728, and for
n
=
1, we get 11
3
=
1331.
Problem 2.2.6.
Find all positive integers n less than
1999
such that n
2
is equal to
the cube of the sum of n’s digits.
(1999 Iberoamerican Mathematical Olympiad)
Solution.
In order for
n
2
to be a cube,
n
must be a cube itself. Because
n
<
1000
we must have
n
=
1
3
,
2
3
, . . .
, or 9
3
. Quick checks show that
n
=
1 and
n
=
27
work, while
n
=
8
,
64, and 125 don’t. As for
n
≥
6
3
=
216, we have
n
2
≥
216
2
>
27
2
. However, the sum of
n
’s digits is at most 9
+
9
+
9
=
27, implying
that no
n
≥
6
3
has the desired property. Thus
n
=
1
,
27 are the only answers.
254
II Solutions, 2. Powers of Integers
Problem 2.2.7.
Prove that for any nonnegative integer n, the number
A
=
2
n
+
3
n
+
5
n
+
6
n
is not a perfect cube.
Solution.
We will use modular arithmetic. A perfect cube has the form 7
k
, 7
k
+
1,
or 7
k
−
1, since
(
7
x
+
1
)
3
≡
(
7
x
+
2
)
3
=
(
7
x
+
4
)
3
≡
1
(
mod 7
)
and
(
7
x
+
3
)
3
≡
(
7
x
+
5
)
3
≡
(
7
x
+
6
)
3
≡ −
1
(
mod 7
).
Now observe that
2
6
=
4
3
≡
1
(
mod 7
),
3
6
=
9
3
≡
2
3
≡
1
(
mod 7
),
5
6
=
(
−
2
)
6
=
2
6
≡
1
(
mod 7
),
6
6
≡
(
−
1
)
6
≡
1
(
mod 7
).
It follows that 2
6
k
≡
3
6
k
≡
5
7
k
≡
6
6
k
≡
1
(
mod 7
)
.
Let
a
n
=
2
n
+
3
n
+
5
n
+
6
n
for
n
≥
0. Set
n
=
6
k
+
r
, with
r
∈ {
0
,
1
,
2
,
3
,
4,
5, 6
}
. Since 2
n
≡
2
r
(
mod 7
)
, 3
n
≡
3
r
(
mod 7
)
, 5
n
≡
5
r
(
mod 7
)
, and 6
n
≡
6
r
(
mod 7
)
, we have
a
n
≡
a
r
(
mod 7
)
.
It is easy to observe that
a
0
≡
a
2
≡
a
6
≡
4
(
mod 7
)
,
a
1
≡
a
4
≡
2
(
mod 7
)
,
and
a
3
≡
5
(
mod 7
)
. Therefore,
a
n
is not a perfect cube.
Problem 2.2.8.
Prove that every integer is a sum of five cubes.
Solution.
For any integer
n
we have the identity
6
n
=
(
n
+
1
)
3
+
(
n
−
1
)
3
+
(
−
n
)
3
+
(
−
n
)
3
.
(
1
)
For an arbitrary integer
m
we choose the integer
v
such that
v
3
≡
m
(
mod 6
)
.
It follows that
m
−
v
3
=
6
n
for some integer
n
and we apply identity (1).
The actual representations are given by (1) and
6
n
+
1
=
6
n
+
1
3
,
6
n
+
2
=
6
(
n
−
1
)
+
2
3
,
6
n
+
3
=
6
(
n
−
4
)
+
3
3
,
6
n
+
4
=
6
(
n
+
1
)
+
(
−
2
)
3
,
6
n
+
5
=
6
(
n
+
1
)
+
(
−
1
)
3
.
255
Remark.
A direct solution is given by the representation
m
=
m
3
+
1
m
+
1
3
+
1
2
3
+
1
m
+
1
3
−
1
2
3
+
−
m
+
1
3
3
+
−
m
+
1
3
3
Problem 2.2.9.
Show that every rational number can be written as a sum of three
cubes.
Solution.
Let
x
be a rational number. We would be done if we could find a relation
of the form
a
3
(
x
)
+
b
3
(
x
)
+
c
3
(
x
)
=
x
, where
a
,
b
,
c
are rational functions. To
make the arithmetic easier, it will actually be convenient to look for a relation
a
3
(
x
)
+
b
3
(
x
)
+
c
3
(
x
)
=
nx
for some integer
n
. Rewrite this as
a
3
(
x
)
+
b
3
(
x
)
=
nx
−
c
3
(
x
)
. Writing
a
(
x
)
=
f
(
x
)/
h
(
x
)
,
b
(
x
)
=
g
(
x
)/
h
(
x
)
for polynomials
f
,
g
,
h
and clearing denominators gives
f
3
(
x
)
+
g
3
(
x
)
=
(
nx
−
c
3
(
x
))
h
3
(
x
).
To build such an equation let
ε
=
cos
2
π
3
+
i
sin
2
π
3
. Then we can write
f
3
(
x
)
+
g
3
(
x
)
=
(
f
(
x
)
+
g
(
x
))(
f
(
x
)
+
ε
g
(
x
))(
f
(
x
)
+
ε
2
g
(
x
)).
It would be convenient if two of the factors on the right were cubes. Then
we could combine them into
h
and we could choose
c
so that the third factor is
nx
−
c
3
(
x
)
. Since we want
f
and
g
to be real, we try
f
(
x
)
+
ε
g
(
x
)
=
(
u
+
εv)
3
f
(
x
)
+
ε
2
g
(
x
)
=
(
u
+
ε
2
v)
3
.
Solving this system (using
ε
2
= −
1
−
ε
) gives
f
(
x
)
=
u
3
−
3
u
v
2
+
v
3
,
g
=
3
u
2
v
−
3
u
v
2
, and hence we are left with solving
nx
−
c
3
(
x
)
=
f
(
x
)
+
g
(
x
)
=
u
3
+
3
u
2
v
−
6
u
v
2
+
v
3
.
Notice that the right-hand side is
(
u
+
v)
3
−
9
u
v
2
. Thus we can take
u
=
x
,
v
=
1,
n
= −
9, and
c
(
x
)
=
u
+
v
=
x
+
1. Solving back through the calculation
gives
f
(
x
)
=
x
3
−
3
x
+
1,
g
(
x
)
=
3
x
2
−
3
x
, and
h
(
x
)
=
x
2
−
x
+
1. Hence we
get
x
3
−
3
x
+
1
x
2
−
x
+
1
2
+
3
x
2
−
3
x
x
2
−
x
+
1
3
+
(
x
+
1
)
3
= −
9
x
,
and the desired conclusion follows by applying this for
x
equal to the desired
rational number divided by
−
9.
Remark.
There are rational numbers that are not the sum of two cubes. We sug-
gest to the reader to find a such example.
256
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