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3.1. General Problems
63
(a)
x
∈
(
−∞
,
−
1
)
. Then
x
≤ −
2 and
x
x
>
2, a contradiction.
(b)
x
= −
1
⇒
x
= −
1. Then
x
x
=
(
−
1
)
·
(
−
1
)
=
1 and
x
x
=
1,
so
x
= −
1 is a solution.
(c)
x
∈
(
−
1
,
0
)
. We have
x
= −
1 and
x
x
= −
x
<
1, false.
(d) If
x
∈ [
0
,
1
)
, then
x
=
0 and
x
x
=
0
<
1, so we have no solution in
this case.
(e) For
x
∈ [
1
,
2
)
we obtain
x
=
1 and
x
x
=
x
, as needed.
(f) Finally, for
x
≥
2 we have
x
≥
2 and
x
x
≥
2
x
≥
4, a contradiction
with (1).
Consequently,
x
∈ {−
1
} ∪ [
1
,
2
)
.
Problem 3.1.3.
Prove that for any integer n one can find integers a and b such
that
n
=
a
√
2
+
b
√
3
.
Solution.
For any integer
n
, one can find an integer
b
such that
b
−
1
<
n
−
√
2
√
3
<
b
.
Because
b
−
2
√
3
<
b
−
1 we obtain
√
2
+
b
√
3
−
2
<
n
≤
√
2
+
b
√
3
.
Using property (5) we have to consider the following cases:
(1) If
n
=
√
2
+
b
√
3
, we are done.
(2) If
n
=
√
2
+
b
√
3
+
1, then
n
=
2
√
2
+
b
√
3
.
(3) If
n
=
√
2
+
b
√
3
−
1, then
n
=
0
√
2
+
b
√
3
.
Problem 3.1.4.
Find all real numbers x
>
1
, such that
n
√
x
n
is an integer for
all positive integers n, n
≥
2
.
(2004 Romanian Regional Mathematical Contest)
Solution.
Put
n
√
x
n
=
a
n
. Then
x
n
=
a
n
n
and
a
n
n
≤
x
n
<
a
n
n
+
1. Taking roots,
one obtains
a
n
≤
x
<
n
a
n
n
+
1. This shows that
x
=
a
n
.
We will show that positive integers
x
,
x
≥
2, satisfy the condition and that they
are the only solutions. Assume, by way of contradiction, that there is a solution
x
that is not a nonnegative integer. Put
x
=
a
+
α
,
a
∈
Z
,
a
≥
1, 0
< α <
1.
It follows that
a
n
< (
a
+
α)
n
<
a
n
+
1, and therefore,
1
<
1
+
α
a
n
<
1
+
1
a
n
≤
2
.
On the other hand, by the Bernoulli inequality,
1
+
α
a
n
≥
1
+
n
α
a
>
2
,
for sufficiently large
n
, a contradiction.
64
I Fundamentals, 3. Floor Function and Fractional Part
Problem 3.1.5.
Let p be a prime and let
α
be a positive real number such that
p
α
2
<
1
4
. Prove that
#
n
√
p
−
α
n
$
=
#
n
√
p
+
α
n
$
for all integers n
≥
#
α/
1
−
2
α
√
p
$
+
1
.
Solution.
It suffices to prove that there are no integers in the interval
n
√
p
−
α
n
,
n
√
p
+
α
n
for
n
> α/
1
−
2
α
√
p
.
Assume by way of contradiction that there is integer
k
such that
n
√
p
−
α
n
<
k
≤
n
√
p
+
α
n
.
Hence
n
2
p
+
α
2
n
2
−
2
α
√
p
<
k
2
≤
n
2
p
+
α
2
n
2
+
2
α
√
p
.
Observe that
α
2
n
2
−
2
α
√
p
>
−
1. If
n
> α/
1
−
2
α
√
p
, then
α
2
n
2
+
2
α
√
p
<
1,
so
n
2
p
−
1
<
k
2
<
n
2
p
+
1
.
It follows that
k
2
=
pn
2
or
√
p
=
k
/
n
, which is false, since
p
is prime.
Problem 3.1.6.
Find the number of different terms of the finite sequence
k
2
1998
,
where k
=
1
,
2
, . . . ,
1997
.
(1998 Balkan Mathematical Olympiad)
Solution.
Note that
'
998
2
1998
(
=
498
<
499
=
'
999
2
1998
(
,
so we can compute the total number of distinct terms by considering
k
=
1
, . . .
,
998 and
k
=
999
, . . . ,
1997 independently. Observe that for
k
=
1
, . . . ,
997,
(
k
+
1
)
2
1998
−
k
2
1998
=
2
k
+
1
1998
<
1
,
so for
k
=
1
, . . . ,
998, each of the numbers
'
1
2
1998
(
=
0
,
1
, . . . ,
498
=
'
998
2
1998
(
appears at least once in the sequence
k
2
/
1998
, for a total of 499 distinct terms.
For
k
=
999
, . . . ,
1996, we have
(
k
+
1
)
2
1998
−
k
2
1998
=
2
k
+
1
1998
>
1
,
3.1. General Problems
65
so the numbers
k
2
/
1998
(
k
=
999
, . . . ,
1997
)
are all distinct, giving
1997
−
999
+
1
=
999 more terms. Thus the total number of distinct terms is
1498.
Problem 3.1.7.
Determine the number of real solutions a of the equation
#
a
2
$
+
#
a
3
$
+
#
a
5
$
=
a
.
(1998 Canadian Mathematical Olympiad)
Solution.
There are 30 solutions. Since
a
/
2
,
a
/
3
, and
a
/
5
are integers, so
is
a
. Now write
a
=
30
p
+
q
for integers
p
and
q
, 0
≤
q
<
30. Then
#
a
2
$
+
#
a
3
$
+
#
a
5
$
=
a
⇔
31
p
+
#
q
2
$
+
#
q
3
$
+
#
q
5
$
=
30
p
+
q
⇔
p
=
q
−
#
q
2
$
−
#
q
3
$
−
#
q
5
$
.
Thus, for each value of
q
, there is exactly one value of
p
(and one value of
a
)
satisfying the equation. Since
q
can equal any of thirty values, there are exactly
30 solutions, as claimed.
Problem 3.1.8.
Let
λ
be the positive root of the equation t
2
−
1998
t
−
1
=
0
.
Define the sequence x
0
,
x
1
, . . .
by setting
x
0
=
1
,
x
n
+
1
=
λ
x
n
,
n
≥
0
.
Find the remainder when x
1998
is divided by
1998
.
(1998 Iberoamerican Mathematical Olympiad)
Solution.
We have
1998
< λ
=
1998
+
√
1998
2
+
4
2
=
999
+
999
2
+
1
<
1999
,
x
1
=
1998,
x
2
=
1998
2
. Since
λ
2
−
1998
λ
−
1
=
0,
λ
=
1998
+
1
λ
and
x
λ
=
1998
x
+
x
λ
for all real numbers
x
. Since
x
n
=
x
n
−
1
λ
and
x
n
−
1
is an integer and
λ
is irra-
tional, we have
x
n
<
x
n
−
1
λ <
x
n
+
1 or
x
n
λ
<
x
n
−
1
<
x
n
+
1
λ
.
66
I Fundamentals, 3. Floor Function and Fractional Part
Since
λ >
1998,
x
n
/λ
=
x
n
−
1
−
1. Therefore,
x
n
+
1
=
x
n
λ
=
#
1998
x
n
+
x
n
λ
$
=
1998
x
n
+
x
n
−
1
−
1
;
hence
x
n
+
1
≡
x
n
−
1
−
1
(
mod 1998
)
. Therefore by induction,
x
1998
≡
x
0
−
999
≡
1000
(
mod 1998
)
.
Problem 3.1.9.
(Hermite
1
)
Let n be a positive integer. Prove that for any real
number x,
nx
=
x
+
%
x
+
1
n
&
+ · · · +
%
x
+
n
−
1
n
&
.
Solution.
Let
f
(
x
)
be the difference between the right-hand side and the left-hand
side of (1). Then
f
x
+
1
n
=
%
x
+
1
n
&
+ · · · +
%
x
+
1
n
+
n
−
1
n
&
−
%
n
x
+
1
n
&
=
%
x
+
1
n
&
+ · · · +
%
x
+
n
−
1
n
&
+
x
+
1
−
nx
+
1
,
and since
x
+
k
=
x
+
k
for each integer
k
, it follows that
f
x
+
1
n
=
f
(
x
)
for all real
x
. Hence
f
is periodic with period 1
/
n
. Thus it suffices to study
f
(
x
)
for 0
≤
x
<
1
/
n
. But
f
(
x
)
=
0 for all these values; hence
f
(
x
)
=
0 for all real
x
, and the proof is complete.
Additional Problems
Problem 3.1.10.
Let
n
be a positive integer. Find with proof a closed formula for
the sum
%
n
+
1
2
&
+
%
n
+
2
2
2
&
+ · · · +
%
n
+
2
k
2
k
+
1
&
+ · · ·
.
(10th International Mathematical Olympiad)
Problem 3.1.11.
Compute the sum
0
≤
i
<
j
≤
n
%
x
+
i
j
&
,
where
x
is a real number.
1
Charles Hermite (1822–1901), French mathematician who did brilliant work in many branches of
mathematics.
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