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8.3
Nonstandard Diophantine Equations
8.3.1
Cubic Equations
Problem 8.3.1.
Find all pairs
(
x
,
y
)
of nonnegative integers such that x
3
+
8
x
2
−
6
x
+
8
=
y
3
.
(1995 German Mathematical Olympiad)
Solution.
Note that for all real
x
,
0
<
5
x
2
−
9
x
+
7
=
(
x
3
+
8
x
2
−
6
x
+
8
)
−
(
x
+
1
)
3
.
Therefore if
(
x
,
y
)
is a solution, we must have
y
≥
x
+
2. In the same vein,
we note that for
x
≥
1,
0
>
−
x
2
−
33
x
+
15
=
(
x
3
+
8
x
2
−
6
x
+
8
)
−
(
x
3
+
9
x
2
+
27
x
+
27
).
Hence we have either
x
=
0, in which case
y
=
2 is a solution, or
x
≥
1, in
which case we must have
y
=
x
+
2. But this means that
0
=
(
x
3
+
8
x
2
−
6
x
+
8
)
−
(
x
3
+
6
x
2
+
12
x
+
8
)
=
2
x
2
−
18
x
.
Hence the only solutions are
(
0
,
2
), (
9
,
11
)
.
Problem 8.3.2.
Find all pairs
(
x
,
y
)
of integers such that
x
3
=
y
3
+
2
y
2
+
1
.
(1999 Bulgarian Mathematical Olympiad)
Solution.
When
y
2
+
3
y
>
0,
(
y
+
1
)
3
>
x
3
>
y
3
. Thus we must have
y
2
+
3
y
≤
0, and
y
= −
3,
−
2,
−
1, or 0, yielding the solutions
(
x
,
y
)
=
(
1
,
0
)
,
(
1
,
−
2
)
, and
(
−
2
,
−
3
)
.
Problem 8.3.3.
Find all the triples
(
x
,
y
,
z
)
of positive integers such that
x y
+
yz
+
zx
−
x yz
=
2
.
First solution.
Let
x
≤
y
≤
z
. We consider the following cases:
160
I Fundamentals, 8. Diophantine Equations
(1) For
x
=
1, we obtain
y
+
z
=
2, and then
(
x
,
y
,
z
)
=
(
1
,
1
,
1
).
(2) If
x
=
2, then 2
y
+
2
z
−
yz
=
2, which gives
(
z
−
2
)(
y
−
2
)
=
2.
The solutions are
z
=
4,
y
=
3 and
z
=
3,
y
=
4. Due to the symmetry of the
relations, the solutions
(
x
,
y
,
z
)
are
(
2
,
3
,
4
), (
2
,
4
,
3
), (
3
,
2
,
4
), (
4
,
2
,
3
), (
3
,
4
,
2
), (
4
,
3
,
2
).
(3) If
x
≥
3,
y
≥
3,
z
≥
3, then
x yz
≥
3
yz
,
x yz
≥
3
x z
,
x yz
≥
3
x y
. Thus
x y
+
x z
+
yz
−
x yz
≤
0, so there are no solutions.
Second solution.
Let
x
=
x
−
1,
y
=
y
−
1,
z
=
z
−
1. The equation is equivalent
to
x
y
z
=
x
+
y
+
z
. If
x
=
0, then
y
=
z
=
0, and we get the solution
(
x
,
y
,
z
)
=
(
1
,
1
,
1
)
for the initial equation. If
x
=
0,
y
=
0, and
x
=
0, then
1
x
y
+
1
y
z
+
1
z
x
=
1
forces one of
x
y
or the other two to be at most 3.
It follows that
(
x
,
y
,
z
)
=
(
1
,
2
,
3
)
and all corresponding permutations and
we get all solutions in case (2).
Problem 8.3.4.
Determine a positive constant c such that the equation
x y
2
−
y
2
−
x
+
y
=
c
has exactly three solutions
(
x
,
y
)
in positive integers.
(1999 United Kingdom Mathematical Olympiad)
Solution.
When
y
=
1 the left-hand side is 0. Thus we can rewrite our equation
as
x
=
y
(
y
−
1
)
+
c
(
y
+
1
)(
y
−
1
)
.
Note that from the offset equation
y
−
1
|
c
and writing
c
=
(
y
−
1
)
d
we
get
x
=
y
+
d
y
+
1
. Hence
d
≡
1
(
mod
y
+
1
)
, and thus
c
≡
y
−
1
(
mod
y
2
−
1
)
.
Conversely, any such
c
makes
x
an integer.
Thus we want
c
to satisfy exactly three congruences
c
≡
y
−
1
(
mod
y
2
−
1
)
.
Every
c
always satisfies this congruence for
y
=
c
+
1, so we need two others.
The first two nontrivial congruences for
y
=
2
,
3 give
c
≡
1
(
mod 3
)
and
c
≡
2
(
mod 8
)
. Hence
c
=
10 is the least solution to both these congruences and also
works for
y
=
11. It does not satisfy any others, since we would have
y
−
1
|
10;
hence
y
=
2
,
3
,
6
,
11. We have already seen that 2, 3, 11 all work, but trying
y
=
6 gives
x
=
2
/
7. Thus there are exactly three solutions with
c
=
10, namely
(
x
,
y
)
=
(
4
,
2
)
,
(
2
,
3
)
, and
(
1
,
11
)
.
8.3. Nonstandard Diophantine Equations
161
Additional Problems
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)
Problem 8.3.6.
Find all positive integers
a
,
b
,
c
such that
a
3
+
b
3
+
c
3
=
2001
.
(2001 Junior Balkan Mathematical Olympiad)
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)
8.3.2
High-Order Polynomial Equations
Problem 8.3.8.
Prove that there are no integers x
,
y
,
z such that
x
4
+
y
4
+
z
4
−
2
x
2
y
2
−
2
y
2
z
2
−
2
z
2
x
2
=
2000
.
Solution.
Suppose by way of contradiction that such numbers exist. Assume with-
out loss of generality that
x
,
y
,
z
are nonnegative integers.
First we prove that the numbers are distinct. For this, consider that
y
=
z
.
Then
x
4
−
4
x
2
y
2
=
2000; hence
x
is even.
Setting
x
=
2
t
yields
t
2
(
t
2
−
y
2
)
=
125. It follows that
t
2
=
25 and
y
2
=
20,
a contradiction.
Let now
x
>
y
>
z
. Since
x
4
+
y
4
+
z
4
is even, at least one of the numbers
x
,
y
,
z
is even and the other two have the same parity. Observe that
x
4
+
y
4
+
z
4
−
2
x
2
y
2
−
2
y
2
z
2
−
2
z
2
x
2
=
(
x
2
−
y
2
)
2
−
2
(
x
2
−
y
2
)
z
2
+
z
4
−
4
y
2
z
2
=
(
x
2
−
y
2
−
z
2
−
2
yz
)(
x
2
−
y
2
−
z
2
+
2
yz
)
=
(
x
+
y
+
z
)(
x
−
y
−
z
)(
x
−
y
+
z
)(
x
+
y
−
z
),
each of the four factors being even. Since 2000
=
16
·
125
=
2
4
·
125, we deduce
that each factor is divisible by 2, but not by 4. Moreover, the factors are distinct:
x
+
y
+
z
>
x
+
y
−
z
>
x
−
y
+
z
>
x
−
y
−
z
.
The smallest even divisors of 2000 that are not divisible by 4 are 2, 10, 50,
250. But 2
·
10
·
50
·
250
>
2000, a contradiction.
162
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