|
II Solutions, 8. Diophantine Equations
A quick check also shows that each
d
n
is odd. Thus because there are infinitely
many solutions
(
b
n
,
d
n
)
to the Pell’s equation (and with each
d
n
odd), there are
infinitely many integral solutions
(
x
n
,
y
n
,
z
n
,
t
n
)
=
10
−
b
n
,
10
+
b
n
,
−
1
2
−
d
n
2
,
−
1
2
+
d
n
2
to the original equation.
8.2.3
Other Quadratic Equations
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)
Solution.
Let
u
=
2
x
+
3,
v
=
2
y
+
3,
w
=
2
z
+
3. Then the given equation is
equivalent to
u
2
+
v
2
+
w
2
=
7
.
It is equivalent to show that the equation
x
2
+
y
2
+
z
2
=
7
w
2
has no nonzero solutions in integers; assume to the contrary that
(
x
,
y
,
z
, w)
is a nonzero solution with
|
w
| + |
x
| + |
y
| + |
z
|
minimal. Modulo 8, we have
x
2
+
y
2
+
z
2
≡
7
w
2
, but every perfect square is congruent to 0, 1, or 4 mod-
ulo 8. Thus we must have
x
,
y
,
z
, w
even, and
(
x
/
2
,
y
/
2
,
z
/
2
, w/
2
)
is a smaller
solution, contradiction.
Remark.
Try to prove the following theorem of Davenport and Cassels:
for n
∈
Z
,
the equation x
2
+
y
2
+
z
2
=
n has rational solutions if and only if it has integer
solutions
. There is a beautiful elementary geometric proof. Try to find it!
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)
Solution.
The only solution is
(
0
,
0
,
0
)
.
Assume, for the sake of contradiction, that there is a triple of integers
(
x
,
y
,
z
)
=
(
0
,
0
,
0
)
satisfying the given equation, and let
(
x
,
y
,
z
)
=
(
x
0
,
y
0
,
z
0
)
be a
nonzero solution that minimizes
|
x
| + |
y
| + |
z
|
>
0.
Because 5
x
2
0
−
14
y
2
0
=
11
z
2
0
, we have
−
2
x
2
0
≡
4
z
2
0
(
mod 7
),
8.2. Quadratic Diophantine Equations
319
or
x
2
0
≡ −
2
z
2
0
≡
5
z
2
0
(
mod 7
)
. Therefore, we have
z
0
≡
0
(
mod 7
)
, because
otherwise we have
5
≡
(
x
0
z
−
1
0
)
2
(
mod 7
),
which is impossible because 5 is not a square modulo 7. (The squares modulo 7
are 0, 1, 2, and 4.)
It follows that
x
0
and
z
0
are divisible by 7, so that 14
y
2
=
5
x
2
−
11
z
2
is
divisible by 49. Therefore, 7
|
y
0
. Then
x
0
7
,
y
0
7
,
z
0
7
is also a solution, but
x
0
7
+
y
0
7
+
z
0
7
<
|
x
0
| + |
y
0
| + |
z
0
|
, contradicting the minimality of
(
x
0
,
y
0
,
z
0
)
.
Therefore, our original assumption was false, and the only integer solution is
(
0
,
0
,
0
)
.
Remark.
A solution mod 8 also works. If
x
or
z
is even, then so is the other, and
hence
y
is even. Thus we can cancel a 2 and get a smaller solution. Suppose we
have a solution with
x
and
z
odd; then we get 5
−
(
0 or 6
)
≡
3
(
mod 8
)
, which
cannot occur.
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)
Solution.
For
n
=
0, we have 2
2
+
1
2
+
1
2
+
1
2
=
7; hence
(
a
,
b
,
c
,
d
)
=
(
2
,
1
,
1
,
1
)
and all permutations. If
n
≥
1, then
a
2
+
b
2
+
c
2
+
d
2
≡
0
(
mod 4
)
;
hence the numbers have the same parity. We analyze two cases.
(a) The numbers
a
,
b
,
c
,
d
are odd. We write
a
=
2
a
+
1, etc. We obtain
4
a
(
a
+
1
)
+
4
b
(
b
+
1
)
+
4
c
(
c
+
1
)
+
4
d
(
d
+
1
)
=
4
(
7
·
4
n
−
1
−
1
).
The left-hand side of the equality is divisible by 8; hence 7
·
4
n
−
1
−
1 must be
even. This happens only for
n
=
1. We obtain
a
2
+
b
2
+
c
2
+
d
2
=
28, with the
solutions
(
3
,
3
,
3
,
1
)
and
(
1
,
1
,
1
,
5
)
.
(b) The numbers
a
,
b
,
c
,
d
are even. Write
a
=
2
a
, etc. We obtain
a
2
+
b
2
+
c
2
+
d
2
=
7
·
4
n
−
1
,
so we proceed recursively.
Finally, we obtain the solutions
(
2
n
+
1
,
2
n
,
2
n
,
2
n
)
,
(
3
·
2
n
−
1
,
3
·
2
n
−
1
,
3
·
2
n
−
1
,
2
n
−
1
)
,
(
2
n
−
1
,
2
n
−
1
,
2
n
−
1
,
5
·
2
n
−
1
)
, and the respective permutations.
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)
320
II Solutions, 8. Diophantine Equations
Solution.
The solutions are
(
0
,
0
,
0
,
2
1002
)
and
(
2
1001
,
2
1001
,
2
1001
,
2
1001
)
.
In order to prove the statement, let
(
x
,
y
,
z
,
t
)
be a solution. Observe that for
odd
a
we have
a
=
4
n
±
1, and
a
2
gives the remainder 1 when divided by 8.
Since the right-hand side is 0
(
mod 8
)
, the equation has no solution with an odd
component.
We thus must have
x
=
2
x
1
,
y
=
2
y
1
,
z
=
2
z
1
,
t
=
2
t
1
, where 0
≤
x
1
≤
y
1
≤
z
1
≤
t
1
are integers and
x
2
1
+
y
2
1
+
z
2
1
+
t
2
1
=
2
2002
. By the same argument,
x
1
=
2
x
2
,
y
1
=
2
y
2
,
z
1
=
2
z
2
,
t
1
=
2
t
2
, where 0
≤
x
2
≤
y
2
≤
z
2
≤
t
2
are
integers and
x
2
2
+
y
2
2
+
z
2
2
+
t
2
2
=
2
2000
.
We can proceed recursively as long as the right-hand side is zero mod 8. Even-
tually we will arrive at
x
=
2
2001
a
,
y
=
2
2001
b
,
z
=
2
2001
c
,
t
=
2
2001
d
, where
0
≤
a
≤
b
≤
c
≤
d
are integers and
a
2
+
b
2
+
c
2
+
d
2
=
4. The only solutions
to this are
(
1
,
1
,
1
,
1
)
and
(
0
,
0
,
0
,
2
)
and the conclusion follows.
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.
Solution.
Suppose
x
=
a
b
,
y
=
c
d
satisfies the given equation, where gcd
(
a
,
b
)
=
gcd
(
c
,
d
)
=
1. Clearing denominators,
(
a
2
+
b
2
)
cd
+
(
c
2
+
d
2
)
ab
=
3
nabcd
.
Thus,
ab
|
(
a
2
+
b
2
)
cd
and
cd
|
(
c
2
+
d
2
)
ab
. Now gcd
(
a
,
b
)
=
1 implies
gcd
(
a
,
a
2
+
b
2
)
=
gcd
(
a
,
b
2
)
=
1, so
ab
|
cd
; likewise,
cd
|
ab
, and together
these give
ab
=
cd
. Thus,
a
2
+
b
2
+
c
2
+
d
2
=
3
nab
.
Now each square on the left is congruent to either 0 or 1 modulo 3. Hence,
either all terms are divisible by 3 or exactly one is. The first case is impossible by
the assumption gcd
(
a
,
b
)
=
gcd
(
c
,
d
)
=
1, and the second is impossible because
ab
=
cd
.
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