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I Fundamentals, 8. Diophantine Equations
Corollary 8.2.2.
The general integral solution to (1) is given by
x
=
k
(
m
2
−
n
2
),
y
=
2
kmn
,
z
=
k
(
m
2
+
n
2
),
(
3
)
where k
,
m
,
n
∈
Z
.
Problem 8.2.1.
Solve the following equation in positive integers:
x
2
+
y
2
=
1997
(
x
−
y
).
(1998 Bulgarian Mathematical Olympiad)
Solution.
The solutions are
(
x
,
y
)
=
(
170
,
145
)
or
(
1827
,
145
).
We have
x
2
+
y
2
=
1997
(
x
−
y
),
2
(
x
2
+
y
2
)
=
2
·
1997
(
x
−
y
),
x
2
+
y
2
+
(
x
2
+
y
2
−
2
·
1997
(
x
−
y
))
=
0
,
(
x
+
y
)
2
+
((
x
−
y
)
2
−
2
·
1997
(
x
−
y
))
=
0
,
(
x
+
y
)
2
+
(
1997
−
x
+
y
)
2
=
1997
2
.
Since
x
and
y
are positive integers, 0
<
x
+
y
<
1997 and 0
<
1997
−
x
+
y
<
1997. Thus the problem reduces to solving
a
2
+
b
2
=
1997
2
in positive integers.
Since 1997 is a prime, gcd
(
a
,
b
)
=
1. By Pythagorean substitution, there are
positive integers
m
>
n
such that gcd
(
m
,
n
)
=
1 and
1997
=
m
2
+
n
2
,
a
=
2
mn
,
b
=
m
2
−
n
2
.
Since
m
2
,
n
2
≡
0
,
1
,
−
1
(
mod 5
)
and 1997
≡
2
(
mod 5
)
, we have
m
,
n
≡
±
1
(
mod 5
)
. Since
m
2
,
n
2
≡
0
,
1
(
mod 3
)
and 1997
≡
2
(
mod 3
)
, we have
m
,
n
≡ ±
1
(
mod 3
)
. Therefore
m
,
n
≡
1
,
4
,
11
,
14
(
mod 15
)
. Since
m
>
n
,
1997
/
2
≤
m
2
≤
1997. Thus we need to consider only
m
=
34
,
41
,
44. The only
solution is
(
m
,
n
)
=
(
34
,
29
)
. Thus
(
a
,
b
)
=
(
1972
,
315
),
which leads to our two solutions.
Problem 8.2.2.
Let p
,
q
,
r be primes and let n be a positive integer such that
p
n
+
q
n
=
r
2
.
Prove that n
=
1
.
(2004 Romanian Mathematical Olympiad)
8.2. Quadratic Diophantine Equations
151
Solution.
Assume that
n
≥
2 satisfies the relation in the problem. Clearly one of
the primes
p
,
q
, and
r
is equal to 2. If
r
=
2 then
p
n
+
q
n
=
4, false, so assume
that
p
>
q
=
2.
Consider the case that
n
>
1 is odd; we have
(
p
+
2
)(
p
n
−
1
−
2
p
n
−
2
+
2
2
p
n
−
3
− · · · +
2
n
−
1
)
=
r
2
.
Notice that
p
n
−
1
−
2
p
n
−
2
+
2
2
p
n
−
3
− · · · +
2
n
−
1
=
2
n
−
1
+
(
p
−
2
)(
p
n
−
2
+
2
2
p
n
−
4
+ · · · +
1
) >
1
and
p
+
2
>
1 hence both factors are equal to
r
. This can be written as
p
n
+
2
n
=
(
p
+
2
)
2
=
p
2
+
4
p
+
4, which is false for
n
≥
3.
Consider the case that
n
>
1 is even and let
n
=
2
m
. From Theorem 8.2.1 it
follows that
p
m
=
a
2
−
b
2
, 2
m
=
2
ab
and
r
=
a
2
+
b
2
, for some integers
a
,
b
with
(
a
,
b
)
=
1. Therefore,
a
and
b
are powers of 2, so
b
=
1 and
a
=
2
m
−
1
.
This implies
p
m
=
4
m
−
1
−
1
<
4
m
, so
p
must be equal to 3. The equality
3
m
=
4
m
−
1
−
1 fails for
m
=
1 and also for
m
≥
2, since 4
m
−
1
>
3
m
+
1, by
induction.
Consequently
n
=
1. For example, in this case we can take
p
=
23,
q
=
2,
and
r
=
5.
Additional Problems
Problem 8.2.3.
Find all Pythagorean triangles whose areas are numerically equal
to their perimeters.
Problem 8.2.4.
Prove that for every positive integer
n
there is a positive integer
k
such that
k
appears in exactly
n
nontrivial Pythagorean triples.
(American Mathematical Monthly)
Problem 8.2.5.
Find the least perimeter of a right-angled triangle whose sides and
altitude are integers.
(Mathematical Reflections)
8.2.2
Pell’s Equation
A special quadratic equation is
u
2
−
D
v
2
=
1
,
(
1
)
where
D
is a positive integer that is not a perfect square. Equation (1) is called
Pell’s
3
equation
, and it has numerous applications in various fields of mathemat-
3
John Pell (1611–1685), English mathematician best known for
Pell’s equation
, which in fact he
had little to do with.
152
I Fundamentals, 8. Diophantine Equations
ics. We will present an elementary approach to solving this equation, due to La-
grange.
Theorem 8.2.3.
If D is a positive integer that is not a perfect square, then equa-
tion
(1)
has infinitely many solutions in positive integers, and the general solution
is given by
(
u
n
, v
n
)
n
≥
1
,
u
n
+
1
=
u
1
u
n
+
D
v
1
v
n
, v
n
+
1
=
v
1
u
n
+
u
1
v
n
,
(
2
)
where
(
u
1
, v
1
)
is its fundamental solution, i.e., the minimal solution different from
(
1
,
0
)
.
Proof.
First, we will prove that equation (1) has a fundamental solution.
Let
c
1
be an integer greater than 1. We will show that there exist integers
t
1
, w
1
≥
1 such that
t
1
−
w
1
√
D
<
1
c
1
, w
1
≤
c
1
.
Indeed, considering
l
k
= [
k
√
D
+
1
]
,
k
=
0
,
1
, . . . ,
c
1
, yields 0
<
l
k
−
k
√
D
≤
1,
k
=
0
,
1
, . . . ,
c
1
, and since
√
D
is an irrational number, it follows that
l
k
=
l
k
whenever
k
=
k
.
There exist
i
,
j
,
p
∈ {
0
,
1
,
2
, . . . ,
c
1
}
,
i
=
j
,
p
=
0, such that
p
−
1
c
1
<
l
i
−
i
√
D
≤
p
c
1
and
p
−
1
c
1
<
l
j
−
j
√
D
≤
p
c
1
because there are
c
1
intervals of the form
p
−
1
c
1
,
p
c
1
,
p
=
0
,
1
, . . . ,
c
1
, and
c
1
+
1
numbers of the form
l
k
−
k
√
D
,
k
=
0
,
1
, . . . ,
c
1
.
Assume
j
>
i
and note that
l
j
>
l
i
.
From the inequalities above it follows that
|
(
l
k
−
l
i
)
−
(
j
−
i
)
√
D
|
<
1
c
1
, and
setting
|
l
j
−
l
i
| =
t
1
and
|
j
−
i
| =
w
1
yields
|
t
1
−
w
1
√
D
|
<
1
c
1
and
w
1
≤
c
1
.
Multiplying this inequality by
t
1
+
w
1
√
D
<
2
w
1
√
D
+
1 gives
|
t
2
1
−
D
w
2
1
|
<
2
w
1
c
1
√
D
+
1
c
1
<
2
√
D
+
1
.
Choosing a positive integer
c
2
>
c
1
such that
|
t
1
−
w
1
√
D
|
>
1
c
2
, we obtain
positive integers
t
2
, w
2
with the properties
|
t
2
2
−
D
w
2
2
|
<
2
√
D
+
1
and
|
t
1
−
t
2
| + |
w
1
−
w
2
| =
0
.
By continuing this procedure, we obtain a sequence of distinct pairs
(
t
n
, w
n
)
n
≥
1
satisfying the inequalities
|
t
2
n
−
D
w
2
n
|
<
2
√
D
+
1 for all positive integers
n
.
It follows that the interval
(
−
2
√
D
−
1
,
2
√
D
+
1
)
contains a nonzero integer
k
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