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I Fundamentals, 8. Diophantine Equations
can be realized by taking the solutions
(
u
k
, v
k
)
of Pell’s equation
u
2
−
2
v
2
=
1,
where
u
0
=
3,
v
0
=
2, and
u
k
, v
k
are obtained from the identity
(
u
0
+
√
2
v
0
)
k
(
u
0
−
√
2
v
0
)
k
=
(
u
k
+
√
2
v
k
)(
u
k
−
√
2
v
k
)
=
1
.
Second solution.
Consider the following identity:
(
a
+
1
)
4
−
(
a
−
1
)
4
=
8
a
3
+
8
a
,
where
a
is a positive integer. Take
a
=
b
3
, where
b
is an even integer. From the
above identity one obtains
(
b
3
+
1
)
4
=
(
2
b
3
)
3
+
(
2
b
)
3
+ [
(
b
3
−
1
)
2
]
2
.
Since
b
is an even number,
b
3
+
1 and
b
3
−
1 are odd numbers. It follows that
the numbers
x
=
2
b
3
,
y
=
2
b
,
z
=
(
b
3
−
1
)
2
, and
t
=
b
3
+
1 have no common
divisor greater than 1.
Additional Problems
Problem 8.2.6.
Let
p
be a prime number congruent to 3 modulo 4. Consider the
equation
(
p
+
2
)
x
2
−
(
p
+
1
)
y
2
+
px
+
(
p
+
2
)
y
=
1
.
Prove that this equation has infinitely many solutions in positive integers, and
show that if
(
x
,
y
)
=
(
x
0
,
y
0
)
is a solution of the equation in positive integers,
then
p
|
x
0
.
(2001 Bulgarian Mathematical Olympiad)
Problem 8.2.7.
Determine all integers
a
for which the equation
x
2
+
ax y
+
y
2
=
1
has infinitely many distinct integer solutions
(
x
,
y
)
.
(1995 Irish Mathematical Olympiad)
Problem 8.2.8.
Prove that the equation
x
3
+
y
3
+
z
3
+
t
3
=
1999
has infinitely many integral solutions.
(1999 Bulgarian Mathematical Olympiad)
8.2. Quadratic Diophantine Equations
157
8.2.3
Other Quadratic Equations
There are many other general quadratic equations that appear in concrete situa-
tions. Here is an example.
Consider the equation
ax y
+
bx
+
cy
+
d
=
0
,
(
1
)
where
a
is a nonzero integer and
b
,
c
,
d
are integers such that
ad
−
bc
=
0.
Theorem 8.2.4.
If
gcd
(
a
,
b
)
=
gcd
(
a
,
c
)
=
1
, then equation (1) is solvable if and
only if there is a divisor m of ad
−
bc such that a
|
m
−
b or a
|
m
−
c.
Proof.
We can write (1) in the following equivalent form:
(
ax
+
c
)(
ay
+
b
)
=
bc
−
ad
.
(
2
)
If such a divisor
m
exists and
a
|
m
−
c
, then we take
ax
+
c
=
m
and
ay
+
b
=
m
, where
mm
=
bc
−
ad
. In order to have solutions, it suffices to
show that
a
|
m
−
b
. Indeed, the relation
mm
=
bc
−
ad
implies
(
ax
+
c
)
m
=
bc
−
ad
, which is equivalent to
a
(
m
x
+
d
)
= −
c
(
m
−
b
)
. Taking into account
that gcd
(
a
,
c
)
=
1, we get
a
|
m
−
b
.
The converse is clearly true.
Remarks.
In case of solvability, equation (1) has only finitely many solutions.
These solutions depend upon the divisors
m
of
ad
−
bc
.
Example.
Solve the equation
3
x y
+
4
x
+
7
y
+
6
=
0
.
Solution.
We have
ad
−
bc
= −
10, whose integer divisors are
−
10,
−
5,
−
2,
−
1,
1, 2, 5, 10. The conditions in Theorem 8.2.4 are satisfied only for
m
= −
5,
−
2,
1, 10. We obtain the solutions
(
x
,
y
)
=
(
−
4
,
−
2
)
,
(
−
3
,
−
3
)
,
(
−
2
,
2
)
,
(
1
,
−
1
)
,
respectively.
In what follows you can find several nonstandard quadratic equations.
Problem 8.2.9.
For any given positive integer n, determine (as a function of n)
the number of ordered pairs
(
x
,
y
)
of positive integers such that
x
2
−
y
2
=
10
2
·
30
2
n
.
Prove further that the number of such pairs is never a perfect square.
(1999 Hungarian Mathematical Olympiad)
Solution.
Because 10
2
·
30
2
n
is even,
x
and
y
must have the same parity. Then
(
x
,
y
)
is a valid solution if and only if
(
u
, v)
=
x
+
y
2
,
x
−
y
2
is a pair of positive
integers that satisfies
u
> v
and
u
v
=
5
2
·
30
2
n
. Now 5
2
·
30
2
n
=
2
2
n
·
3
2
n
·
5
2
n
+
2
158
I Fundamentals, 8. Diophantine Equations
has exactly
(
2
n
+
1
)
2
(
2
n
+
3
)
factors. Thus without the condition
u
> v
there
are exactly
(
2
n
+
1
)
2
(
2
n
+
3
)
such pairs
(
u
, v)
. Exactly one pair has
u
=
v
, and
by symmetry half of the remaining pairs have
u
> v
. It follows that there are
1
2
((
2
n
+
1
)
2
(
2
n
+
3
)
−
1
)
=
(
n
+
1
)(
4
n
2
+
6
n
+
1
)
valid pairs.
Now suppose that
(
n
+
1
)(
4
n
2
+
6
n
+
1
)
were a square. Because
n
+
1 and
4
n
2
+
6
n
+
1
=
(
4
n
+
2
)(
n
+
1
)
−
1 are coprime, 4
n
2
+
6
n
+
1 must be a square
as well. However,
(
2
n
+
1
)
2
<
4
n
2
+
6
n
+
1
< (
2
n
+
2
)
2
, a contradiction.
Problem 8.2.10.
Prove that the equation a
2
+
b
2
=
c
2
+
3
has infinitely many
integer solutions
{
a
,
b
,
c
}
.
(1996 Italian Mathematical Olympiad)
Solution.
Let
a
be any odd number, let
b
=
(
a
2
−
5
)/
2 and
c
=
(
a
2
−
1
)/
2. Then
c
2
−
b
2
=
(
c
+
b
)(
c
−
b
)
=
a
2
−
3
.
Remark.
Actually one can prove that any integer
n
can be represented in infinitely
many ways in the form
a
2
+
b
2
−
c
2
with
a
,
b
,
c
∈
Z
.
Additional Problems
Problem 8.2.11.
Prove that the equation
x
2
+
y
2
+
z
2
+
3
(
x
+
y
+
z
)
+
5
=
0
has no solutions in rational numbers.
(1997 Bulgarian Mathematical Olympiad)
Problem 8.2.12.
Find all integers
x
,
y
,
z
such that 5
x
2
−
14
y
2
=
11
z
2
.
(2001 Hungarian Mathematical Olympiad)
Problem 8.2.13.
Let
n
be a nonnegative integer. Find the nonnegative integers
a
,
b
,
c
,
d
such that
a
2
+
b
2
+
c
2
+
d
2
=
7
·
4
n
.
(2001 Romanian JBMO Team Selection Test)
Problem 8.2.14.
Prove that the equation
x
2
+
y
2
+
z
2
+
t
2
=
2
2004
,
where 0
≤
x
≤
y
≤
x
≤
t
, has exactly two solutions in the set of integers.
(2004 Romanian Mathematical Olympiad)
8.3. Nonstandard Diophantine Equations
159
Problem 8.2.15.
Let
n
be a positive integer. Prove that the equation
x
+
y
+
1
x
+
1
y
=
3
n
does not have solutions in positive rational numbers.
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