|
Second solution.
We show that 3
N
appears in exactly
N
Pythagorean triples.
Obviously 3
N
cannot be the even side. To see that 3
N
cannot be the hypotenuse,
suppose
N
is the least power of 3 that occurs as the hypotenuse. Squares are 0 or 1
(
mod 3
)
; hence 3
|
x
2
+
y
2
implies 3
|
x
and 3
|
y
. But then canceling a common
factor of 3 gives a Pythagorean triple with 3
N
−
1
as the hypotenuse, contradicting
the choice of
N
. Thus the number of Pythagorean triples in which it appears is
exactly the number of solutions to 3
N
=
k
(
m
2
−
n
2
)
with gcd
(
m
,
n
)
=
1. For
such a solution
k
=
3
r
for some 0
≤
r
≤
N
−
1 and
(
m
−
n
)(
m
+
n
)
=
3
N
−
r
. But
if
m
+
n
and
m
−
n
are both nontrivial powers of 3, then we get 3
|
m
and 3
|
n
,
contradicting the fact that gcd
(
m
,
n
)
=
1. Thus
m
+
n
=
3
N
−
r
and
m
−
n
=
1.
Thus we get exactly
N
solutions.
Problem 8.2.5.
Find the least perimeter of a right-angled triangle whose sides
and altitude are integers.
(Mathematical Reflections)
8.2. Quadratic Diophantine Equations
315
Solution.
The answer for the least possible perimeter is 60. It holds for a right-
angled triangle
(
15
,
20
,
25
)
, whose altitude is 12.
Let
x
,
y
,
z
be a Pythagorean triple with
z
the hypotenuse and let
h
be the
altitude of the triangle. Let
d
=
gcd
(
x
,
y
,
z
)
be the greatest common divisor of
x
,
y
,
z
. We have that
x
=
d
·
a
,
y
=
d
·
b
, and
z
=
d
·
c
for
a
,
b
,
c
a primitive
Pythagorean triple. Now calculating the area of the triangle in two ways, we obtain
that
h
=
x y
z
=
abd
c
. Using the fact that gcd
(
ab
,
c
)
=
1, we get
c
|
d
, which tells
us that
c
≤
d
, since both are positive integers. Because the perimeter is equal to
d
(
a
+
b
+
c
)
≥
c
(
a
+
b
+
c
)
, we can minimize it by taking
c
=
d
. Then the
altitude of a right-angled triangle having sides
ca
,
cb
,
c
2
(with
c
2
the hypotenuse)
is
ab
, an integer. We get the perimeter
p
=
c
(
a
+
b
+
c
).
(
1
)
We know that
a
,
b
,
c
is a primitive Pythagorean triple if and only if there exist
m
,
n
∈
Z
+
such that gcd
(
m
,
n
)
=
1,
m
≡
n
(
mod 2
)
,
m
>
n
>
0, that satisfy
a
=
m
2
−
n
2
,
b
=
2
mn
,
c
=
m
2
+
n
2
. Replacing in (1), we notice that all we
need to find is the minimum value of
p
=
(
m
2
+
n
2
)(
2
m
2
+
2
mn
).
Clearly
m
>
n
>
0; therefore
m
≥
2 and
n
≥
1. Thus
p
≥
(
2
2
+
1
)(
2
·
2
2
+
2
·
2
·
1
)
=
60
.
Now the triangle with sides
(
15
,
20
,
25
)
satisfies all the conditions of the orig-
inal problem, and its perimeter is 60. the problem is solved.
8.2.2
Pell’s Equation
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)
Solution.
We show first that
p
|
x
. Substituting
y
=
z
+
1 and rewriting, we
obtain
x
2
=
(
z
−
x
)((
p
+
1
)(
z
+
x
)
+
p
).
Let
q
=
gcd
(
z
−
x
, (
p
+
1
)(
z
+
x
)
+
p
)
. Then
q
|
x
, and therefore
q
|
z
,
and also
q
|
p
. On the other hand,
q
=
1, because otherwise both factors on the
316
II Solutions, 8. Diophantine Equations
right-hand side must be perfect squares, yet
(
p
+
1
)(
z
+
x
)
+
p
≡
3
(
mod 4
)
.
Thus
q
=
p
and
p
|
x
as desired.
Now write
x
=
px
1
and
z
=
pz
1
to obtain
x
2
1
=
(
z
1
−
x
1
)((
p
+
1
)(
z
1
+
x
1
)
+
1
).
By what we showed above, the two terms on the right are coprime and thus
must be perfect squares. Therefore, for some
a
,
b
we have
z
1
−
x
1
=
a
2
, (
p
+
1
)(
z
1
+
x
1
)
+
1
=
b
2
,
x
1
=
ab
.
The above equality implies
b
2
=
(
p
+
1
)(
a
2
+
2
ab
)
+
1
,
i.e.,
(
p
+
2
)
b
2
−
(
p
+
1
)(
a
+
b
)
2
=
1
.
Conversely, given
a
and
b
satisfying the last equation, there exists a unique
pair
(
x
1
,
y
1
)
satisfying the equation above, and hence a unique pair
(
x
,
y
)
satisfy-
ing the original equation.
Thus, we reduced the original equation to a “Pell-type” equation. To get some
solutions, look at the odd powers of
√
p
+
2
+
√
p
+
1. It follows easily that
(
p
+
2
+
p
+
1
)
2
k
+
1
=
m
k
p
+
2
+
n
k
p
+
1
for some positive integers
m
k
,
n
k
. Then
(
p
+
2
−
p
+
1
)
2
k
+
1
=
m
k
p
+
2
−
n
k
p
+
1
,
and multiplying the left and right sides gives
(
p
+
2
)
m
2
k
−
(
p
+
1
)
n
2
k
=
1
.
Clearly,
n
k
>
m
k
, so setting
b
k
=
m
k
,
a
k
=
n
k
−
m
k
gives a solution for
(
a
,
b
)
.
Finally, it is easy to see that the sequences
{
m
k
}
,
{
n
k
}
are strictly increasing, so
we obtain infinitely many solutions this way.
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)
8.2. Quadratic Diophantine Equations
317
Solution.
The equation has infinitely many solutions if and only if
a
2
≥
4.
Rewrite the given equation in the form
(
2
x
+
ay
)
2
−
(
a
2
−
4
)
y
2
=
4
.
If
a
2
<
4, the real solutions to this equation form an ellipse, and so only
finitely many integer solutions occur. If
a
= ±
2, there are infinitely many solu-
tions, since the equation becomes
(
x
±
y
)
2
=
1. If
a
2
>
4, then
a
2
−
4 is not a
perfect square, and so the Pell’s equation
u
2
−
(
a
2
−
4
)v
2
=
1 has infinitely many
solutions. But setting
x
=
u
−
a
v
,
y
=
2
v
gives infinitely many solutions of the
given equation.
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)
Solution.
Observe that
(
m
−
n
)
3
+
(
m
+
n
)
3
=
2
m
3
+
6
mn
2
. Now suppose we
want a general solution of the form
(
x
,
y
,
z
,
t
)
=
a
−
b
,
a
+
b
,
c
2
−
d
2
,
c
2
+
d
2
for integers
a
,
b
and odd integers
c
,
d
. One simple solution to the given equation
is
(
x
,
y
,
z
,
t
)
=
(
10
,
10
,
−
1
,
0
)
, so we try setting
a
=
10 and
c
= −
1. Then
(
x
,
y
,
z
,
t
)
=
10
−
b
,
10
+
b
,
−
1
2
−
d
2
,
−
1
2
+
d
2
is a solution exactly when
(
2000
+
60
b
2
)
−
1
+
3
d
2
4
=
1999
,
i.e.,
d
2
−
80
b
2
=
1
.
The second equation is a Pell’s equation with solution
(
d
1
,
b
1
)
=
(
9
,
1
)
. We
can generate infinitely many more solutions by setting
(
d
n
+
1
,
b
n
+
1
)
=
(
9
d
n
+
80
b
n
,
9
b
n
+
d
n
)
for
n
=
1
,
2
,
3
, . . .
This can be proved by induction, and it follows from a general recursion
(
p
n
+
1
,
q
n
+
1
)
=
(
p
1
p
n
+
q
1
q
n
D
,
p
1
q
n
+
q
1
p
n
)
for generating solutions to
p
2
−
Dq
2
=
1 given a nontrivial solution
(
p
1
,
q
1
)
.
318
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