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8.3
Nonstandard Diophantine Equations
8.3.1
Cubic Equations
Problem 8.3.5.
Find all triples
(
x
,
y
,
z
)
of natural numbers such that y is a prime
number, y and
3
do not divide z, and x
3
−
y
3
=
z
2
.
(1999 Bulgarian Mathematical Olympiad)
8.3. Nonstandard Diophantine Equations
321
Solution.
We rewrite the equation in the form
(
x
−
y
)(
x
2
+
x y
+
y
2
)
=
z
2
.
Any common divisor of
x
−
y
and
x
2
+
x y
+
y
2
also divides both
z
2
and
(
x
2
+
x y
+
y
2
)
−
(
x
+
2
y
)(
x
−
y
)
=
3
y
2
. Because
z
2
and 3
y
2
are relatively
prime by assumption,
x
−
y
and
x
2
+
x y
+
y
2
must be relatively prime as well.
Therefore, both
x
−
y
and
x
2
+
x y
+
y
2
are perfect squares.
Writing
a
=
√
x
−
y
, we have
x
2
+
x y
+
y
2
=
(
a
2
+
y
)
2
+
(
a
2
+
y
)
y
+
y
2
=
a
4
+
3
a
2
y
+
3
y
2
and
4
(
x
2
+
x y
+
y
2
)
=
(
2
a
2
+
3
y
)
2
+
3
y
2
.
Writing
m
=
2
x
2
+
x y
+
y
2
and
n
=
2
a
2
+
3
y
, we have
m
2
=
n
2
+
3
y
2
,
or
(
m
−
n
)(
m
+
n
)
=
3
y
2
,
so
(
m
−
n
,
m
+
n
)
=
(
1
,
3
y
2
)
,
(
y
,
3
y
)
, or
(
3
,
y
2
)
.
In the first case, 2
n
=
3
y
2
−
1 and 4
a
2
=
2
n
−
6
y
=
3
y
2
−
6
y
−
1. Hence,
a
2
≡
2
(
mod 3
)
, which is impossible.
In the second case,
n
=
y
<
2
a
2
+
3
y
=
n
, a contradiction.
In the third case, write 4
a
2
=
2
n
−
6
y
=
y
2
−
6
y
−
3 as 4
a
2
=
(
y
−
3
)
2
−
12,
and so 12
=
(
y
−
3
)
2
−
a
2
=
(
y
−
a
−
3
)(
y
+
a
−
3
)
. Since these factors are
congruent mod 2, we must have
y
−
a
−
3
=
2 and
y
+
a
+
3
=
6, so
a
=
1,
y
=
7. This yields the unique solution
(
x
,
y
,
z
)
=
(
8
,
7
,
13
)
.
Problem 8.3.6.
Find all the positive integers a
,
b
,
c such that
a
3
+
b
3
+
c
3
=
2001
.
(2001 Junior Balkan Mathematical Olympiad)
Solution.
Assume without loss of generality that
a
≤
b
≤
c
.
It is obvious that 1
3
+
10
3
+
10
3
=
2001. We prove that
(
1
,
10
,
10
)
is the only
solution of the equation, except for its permutations.
We start by proving a useful lemma:
Lemma.
Suppose n is an integer. The remainder of n
3
when divided by
9
is
0
,
1
,
or
−
1
.
Indeed, if
n
=
3
k
, then 9
|
n
3
, and if
n
=
3
k
±
1, then
n
3
=
27
k
3
±
27
k
2
+
9
k
±
1
=
M
9
±
1.
322
II Solutions, 8. Diophantine Equations
Since 2001
=
9
·
222
+
3
=
M
9
+
3, then
a
3
+
b
3
+
c
3
=
2001 implies
a
3
=
M
9
+
1,
b
3
=
M
9
+
1 and
c
3
=
M
9
+
1; hence
a
,
b
,
c
are numbers of the
form
M
3
+
1. We search for
a
,
b
,
c
in the set
{
1
,
4
,
7
,
10
,
13
, . . .
}
.
If
c
≥
13 then
c
3
≥
2197
>
2001
=
a
3
+
b
3
+
c
3
, which is false. If
c
≤
7
then 2001
=
a
3
+
b
3
+
c
3
≤
3
·
343, which again is false. Hence
c
=
10 and
consequently
a
3
+
b
3
=
1001. If
b
<
c
=
10 then
a
≤
b
≤
7 and 1001
=
a
3
+
b
3
≤
2
·
7
3
=
2
·
343, a contradiction. Thus
b
=
10 and
a
=
1.
Therefore
(
a
,
b
,
c
)
∈ {
(
1
,
10
,
10
), (
10
,
1
,
10
), (
10
,
10
,
1
)
}
.
Problem 8.3.7.
Determine all ordered pairs
(
m
,
n
)
of positive integers such that
n
3
+
1
mn
−
1
is an integer.
(35th International Mathematical Olympiad)
First solution.
Let
n
3
+
1
mn
−
1
=
k
,
k
a positive integer.
From
n
3
+
1
=
k
(
mn
−
1
)
, one obtains
k
+
1
=
n
(
km
−
n
2
)
. Thus,
n
divides
k
+
1 and by noting
km
−
n
2
=
q
one has
k
=
nq
−
1. Using this form of
k
we
have
n
3
+
1
=
(
nq
−
1
)(
mn
−
1
)
⇔
n
(
mq
−
n
)
=
m
+
q
.
Since
m
+
q
>
0, it follows that
x
=
mq
−
n
>
0. Thus we have the system
xn
=
m
+
q
,
x
+
n
=
mq
.
By adding these equations we obtain
xn
+
mq
=
x
+
n
+
m
+
q
⇔
xn
+
mq
−
x
−
n
−
m
−
q
+
2
=
2
⇔
(
x
−
1
)(
n
−
1
)
+
(
m
−
1
)(
q
−
1
)
=
2
.
The equation
(
x
−
1
)(
n
−
1
)
+
(
m
−
1
)(
q
−
1
)
=
2
has only a finite number of positive integer solutions. These are listed below:
(1)
x
=
1
,
m
−
1
=
2
,
q
−
1
=
1
⇒
x
=
1
,
m
=
3
,
q
=
2
⇒
m
=
3,
n
=
5.
(2)
x
=
1
,
m
−
1
=
1
,
q
−
1
=
2
⇒
m
=
2
,
n
=
5.
(3)
n
=
1
,
m
−
1
=
2
,
q
−
1
=
1
⇒
n
=
1
,
m
=
3.
8.3. Nonstandard Diophantine Equations
323
(4)
n
=
1
,
m
−
1
=
1
,
q
−
1
=
2
⇒
n
=
1
,
m
=
2.
(5)
m
=
1
,
x
−
1
=
2
,
n
−
1
=
1
⇒
m
=
1
,
n
=
2.
(6)
m
=
1
,
x
−
1
=
1
,
n
−
1
=
2
⇒
m
=
1
,
n
=
3.
(7)
q
=
1
,
x
−
1
=
1
,
n
−
1
=
2
⇒
n
=
3
,
m
=
5.
(8)
q
=
1
,
x
−
1
=
2
,
n
−
1
=
1
⇒
n
=
2
,
m
=
5.
(9)
x
−
1
=
n
−
1
=
m
−
1
=
q
−
1
=
1
⇒
m
=
n
=
2.
Thus, we have obtained the following nine pairs
(
m
,
n
)
:
(
5
,
3
), (
3
,
5
), (
5
,
2
)
,
(
2
,
5
)
,
(
3
,
1
), (
1
,
3
), (
2
,
1
), (
1
,
2
), (
2
,
2
)
. All pairs are solutions of the problem.
Second solution.
Note that
(
m
,
n
)
is a solution if and only if
m
2
n
2
+
mn
+
1
+
n
3
+
1
mn
−
1
=
n
3
(
m
3
+
1
)
mn
−
1
is an integer. Since gcd
(
n
,
mn
−
1
)
=
1, this occurs if and only if
m
3
+
1
mn
−
1
is a
solution. That is,
(
m
,
n
)
is a solution if and only if
(
n
,
m
)
is a solution. Suppose
(
m
,
n
)
is a solution and
mn
>
1; then defining
q
as in the current solution, we
see
(
q
,
n
)
is a solution. Further, since
(
n
2
−
1
)
2
≥
n
3
+
1 for
n
≥
2, we see that
q
<
n
. Similarly, if
m
=
n
>
2, we get
q
<
n
. Thus following the steps
(i) If
m
<
n
, interchange
m
and
n
,
(ii) If
m
>
n
>
1 or
m
=
n
>
2, replace
(
m
,
n
)
by
(
q
,
n
)
,
starting from any solution we can always reduce to a smaller solution until we
get to either
m
=
n
=
2 or
n
=
1. In the latter case we have
m
=
2 or 3.
Backtracking gives the chains of solutions
(
3
,
5
(
→
(
5
,
3
)
→
(
1
,
3
)
→
(
3
,
1
)
,
(
2
,
5
)
→
(
5
,
2
)
→
(
1
,
2
)
→
(
2
,
1
)
and the single solution
(
2
,
2
)
. Thus these nine
pairs are all solutions to the problem.
8.3.2
High-Order Polynomial Equations
Problem 8.3.12.
Prove that there are no positive integers x and y such that
x
5
+
y
5
+
1
=
(
x
+
2
)
5
+
(
y
−
3
)
5
.
Solution.
Notice that
z
5
≡
z
(
mod 5
)
; hence
x
+
y
+
1
≡
(
x
+
2
)
+
(
y
−
3
)
(
mod 5
)
, impossible.
Problem 8.3.13.
Prove that the equation y
2
=
x
5
−
4
has no integer solutions.
(1998 Balkan Mathematical Olympiad)
324
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