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I Fundamentals, 8. Diophantine Equations
Problem 8.3.9.
Find the smallest value for n for which there exist positive integers
x
1
, . . . ,
x
n
with
x
4
1
+
x
4
2
+ · · · +
x
4
n
=
1998
.
Solution.
Observe that for any integer
x
we have
x
4
=
16
k
or
x
4
=
16
k
+
1 for
some
k
.
Since 1998
=
16
·
124
+
14, it follows that
n
≥
14.
If
n
=
14, all the numbers
x
1
,
x
2
, . . . ,
x
14
must be odd, so let
x
4
k
=
16
a
k
+
1.
Then
a
k
=
x
4
k
−
1
16
,
k
=
1
,
2
, . . . ,
14; hence
a
k
∈ {
0
,
5
,
39
,
150
, . . .
}
and
a
1
+
a
2
+
· · · +
a
14
=
124. It follows that
a
k
∈ {
0
,
5
,
39
}
for all
k
=
1
,
2
, . . . ,
14, and
since 124
=
5
·
24
+
4, the number of terms
a
k
equal to 39 is 1 or at least 6. A
simple analysis shows that the claim fails in both cases; hence
n
≥
15. Any of the
equalities
1998
=
5
4
+
5
4
+
3
4
+
3
4
+
3
4
+
3
4
+
3
4
+
3
4
+
3
4
+
3
4
+
3
4
+
2
4
+
1
4
+
1
4
+
1
4
=
5
4
+
5
4
+
4
4
+
3
4
+
3
4
+
3
4
+
3
4
+
3
4
+
3
4
+
1
4
+
1
4
+
1
4
+
1
4
+
1
4
+
1
4
proves that
n
=
15.
Problem 8.3.10.
Find all positive integer solutions
(
x
,
y
,
z
,
t
)
of the equation
(
x
+
y
)(
y
+
z
)(
z
+
x
)
=
t x yz
such that
gcd
(
x
,
y
)
=
gcd
(
y
,
z
)
=
gcd
(
z
,
x
)
=
1
.
(1995 Romanian International Mathematical Olympiad Team Selection Test)
Solution.
It is obvious that
(
x
,
x
+
y
)
=
(
x
,
x
+
z
)
=
1, and
x
divides
y
+
z
,
y
divides
z
+
x
, and
z
divides
x
+
y
. Let
a
,
b
, and
c
be integers such that
x
+
y
=
cz
,
y
+
z
=
ax
,
z
+
x
=
by
.
We may assume that
x
≥
y
≥
z
. If
y
=
z
, then
y
=
z
=
1 and then
x
∈ {
1
,
2
}
. If
x
=
y
, then
x
=
y
=
1 and
z
=
1. So assume that
x
>
y
>
z
. Since
a
=
y
+
z
x
<
2, we have
a
=
1 and
x
=
y
+
z
. Thus,
y
|
y
+
2
z
and
y
|
2
z
. Since
y
>
z
,
y
=
2
z
and since gcd
(
y
,
z
)
=
1, one has
z
=
1,
y
=
2,
x
=
3.
Finally, the solutions are
(
1
,
1
,
1
,
8
)
,
(
2
,
1
,
1
,
9
)
,
(
3
,
2
,
1
,
10
)
and those ob-
tained by permutations of
x
,
y
,
z
.
8.3. Nonstandard Diophantine Equations
163
Problem 8.3.11.
Determine all triples of positive integers a
,
b
,
c such that a
2
+
1
,
b
2
+
1
are prime and
(
a
2
+
1
)(
b
2
+
1
)
=
c
2
+
1
.
(2002 Polish Mathematical Olympiad)
Solution.
Of course, we may assume that
a
≤
b
. Since
a
2
(
b
2
+
1
)
=
(
c
−
b
)(
c
+
b
)
and
b
2
+
1 is a prime, we have
b
2
+
1
|
c
−
b
or
b
2
+
1
|
c
+
b
. If
b
2
+
1
|
c
−
b
,
then
a
2
≥
c
+
b
≥
b
2
+
2
b
+
1; impossible, since
a
≤
b
. So there is
k
such that
c
+
b
=
k
(
b
2
+
1
)
and
a
2
=
k
(
b
2
+
1
)
−
2
b
. Thus,
b
2
≥
k
(
b
2
+
1
)
−
2
b
>
kb
2
−
2
b
,
whence
k
≤
2. If
k
=
2, then
b
2
≥
2
b
2
−
2
b
+
2; thus
(
b
−
1
)
2
+
1
≤
0, false.
Thus
k
=
1 and
a
=
b
−
1. But then
b
2
+
1 and
(
b
−
1
)
2
+
1 are primes and at least
one of them is even, forcing
b
−
1
=
1 and
b
=
2,
a
=
1,
c
=
3. By symmetry,
we obtain
(
a
,
b
,
c
)
=
(
1
,
2
,
3
)
or
(
2
,
1
,
3
)
.
Additional Problems
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
.
Problem 8.3.13.
Prove that the equation
y
2
=
x
5
−
4 has no integer solutions.
(1998 Balkan Mathematical Olympiad)
Problem 8.3.14.
Let
m
,
n
>
1 be integers. Solve in positive integers the equation
x
n
+
y
n
=
2
m
.
(2003 Romanian Mathematical Olympiad)
Problem 8.3.15.
For a given positive integer
m
, find all triples
(
n
,
x
,
y
)
of positive
integers such that
m
,
n
are relatively prime and
(
x
2
+
y
2
)
m
=
(
x y
)
n
, where
n
,
x
,
y
can be represented in terms of
m
.
(1995 Korean Mathematical Olympiad)
8.3.3
Exponential Diophantine Equations
Problem 8.3.16.
Find the integer solutions to the equation
9
x
−
3
x
=
y
4
+
2
y
3
+
y
2
+
2
y
.
Solution.
We have successively
4
((
3
x
)
2
−
3
x
)
+
1
=
4
y
4
+
8
y
3
+
4
y
2
+
8
y
+
1
,
164
I Fundamentals, 8. Diophantine Equations
then
(
2
t
−
1
)
2
=
4
y
4
+
8
y
3
+
4
y
2
+
8
y
+
1
,
where 3
x
=
t
≥
1, since it is clear that there are no solutions with
x
<
0.
Observe that
(
2
y
2
+
2
y
)
2
<
E
≤
(
2
y
2
+
2
y
+
1
)
2
.
Since
E
=
(
2
t
−
1
)
2
is a square, then
E
=
(
2
y
2
+
2
y
+
1
)
2
if and only if
4
y
(
y
−
1
)
=
0
,
so
y
=
0 or
y
=
1.
If
y
=
0, then
t
=
1 and
x
=
0.
If
y
=
1, then
t
=
3 and
x
=
1.
Hence the solutions
(
x
,
y
)
are
(
0
,
0
)
and
(
1
,
1
)
.
Problem 8.3.17.
The positive integers x
,
y
,
z satisfy the equation
2
x
x
=
y
y
+
z
z
.
Prove that x
=
y
=
z.
(1997 St. Petersburg City Mathematical Olympiad)
Solution.
We note that
(
x
+
1
)
x
+
1
≥
x
x
+
1
+
(
x
+
1
)
x
x
>
2
x
x
. Thus we cannot
have
y
>
x
or
z
>
x
, since otherwise, the right side of the equation will exceed
the left. But then 2
x
x
≥
y
y
+
z
z
, with equality if and only if
x
=
y
=
z
.
Problem 8.3.18.
Find all solutions in nonnegative integers x
,
y
,
z of the equation
2
x
+
3
y
=
z
2
.
(1996 United Kingdom Mathematical Olympiad)
Solution.
If
y
=
0, then 2
x
=
z
2
−
1
=
(
z
+
1
)(
z
−
1
)
, so
z
+
1 and
z
−
1 are powers
of 2. The only powers of 2 that differ by 2 are 4 and 2, so
(
x
,
y
,
z
)
=
(
3
,
0
,
3
)
.
If
y
>
0, then taking the equation mod 3, it follows that
x
is even. Now we
have 3
y
=
z
2
−
2
x
=
(
z
+
2
x
/
2
)(
z
−
2
x
/
2
)
. The factors are powers of 3, say
z
+
2
x
/
2
=
3
m
and
z
−
2
x
/
2
=
3
n
, but then 3
m
−
3
n
=
2
x
/
2
+
1
. Since the right side
is not divisible by 3, we must have
n
=
0 and
3
m
−
1
=
2
x
/
2
+
1
.
If
x
=
0, we have
m
=
1, yielding
(
x
,
y
,
z
)
=
(
0
,
1
,
2
)
. Otherwise, 3
m
−
1 is
divisible by 4, so
m
is even and 2
x
/
2
+
1
=
(
3
m
/
2
+
1
)(
3
m
/
2
−
1
)
. The two factors
on the right are powers of 2 differing by 2, so they are 2 and 4, giving
x
=
4 and
(
x
,
y
,
z
)
=
(
4
,
2
,
5
)
.
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