Download Pythagorean Triples and Rational Points on the Unit Circle

Pythagorean Triples and
Rational Points on the Unit Circle
Solutions
Below are sample solutions to the problems posed. You may find that your solutions
are different in form and you may have found patterns not listed here.
Exploring Triangles Algebraically
• Take a moment and list some Pythagorean triples that you know. Are there any
patterns in your list? For instance, in any Pythagorean triple, how many even
entries are possible? Are 0, 1, 2, and 3 all possible? Can the hypotenuse be the
only even side length? Justify your answer.
The key to this problem is that the square of an even number is even
and the square of an odd number is odd. So, we cannot have all of a, b,
and c being odd since if the sum or difference of two odd numbers must
be even. So, 0 even numbers is impossible. Clearly, having all of a, b,
and c as even numbers is possible as (6, 8, 10) is a Pythagorean triple.
Also, a single even is possible since (3, 4, 5) is a triple. Since the sum or
difference of even numbers is also even, it is impossible to have exactly
two numbers in the triple be even. Thus, there can only be triples
with a single even number or three even numbers.
In the case of a single even number in a Pythagorean triple, it is
impossible for the hypotenuse to be even. If a = 2j + 1 and
b = 2k + 1 are the odd triangle lengths and c = 2m is the even
hypotenuse length, then we have
(2j + 1)2 + (2k + 1)2 = (2m)2 =⇒
4j 2 + 4j + 1 + 4k 2 + 4k + 1 = 4m2 =⇒
4(j 2 + j + k 2 + k) + 2 = 4m2
The left-hand side is two more than a multiple of 4 while the
right-hand side is a multiple of 4; this is impossible.
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• Notice that (3, 4, 5) and (6, 8, 10) are two triples that are, in some sense, very
closely related. How? Can you find another triple that is related to (3, 4, 5) and
to (6, 8, 10)? Can you use this to show that there are infinitely many right
triangle triples? Explain why your result is true.
Multiplying a given Pythagorean triple by a positive integer, k yields
another Pythagorean triple. That is, if (a, b, c) is a Pythagorean triple,
then so is (ka, kb, kc) since
(ka)2 + (kb)2 = k 2 (a2 + b2 ) = k 2 c2 = (kc)2 .
Since there are infinitely many positive integers, each Pythagorean
triple generates an infinite family of triples.
• Consider the following table of Pythagorean triples.
a
3
5
7
9
11
13
b
4
12
24
40
60
84
c
5
13
25
41
61
85
Spend some time determining how these triples are related. Can you determine
the next three triples in this pattern? (Be sure to check that your triples are
Pythagorean triples!)
The next three triples are (15, 112, 113), (17, 144, 145) and (19, 180, 181).
It is obvious that c = b + 1 for each of these triples and that a is
running through the odd numbers in order. Then,
a2 + b2 = (b + 1)2 =⇒ a2 = 2b + 1,
1
or b = (a2 − 1).
2
• Use the pattern you found in the previous problem to create a formula for these
triples and will hence generate an infinite number of Pythagorean triples.
Prove that your formula always gives a Pythagorean triple.
From the previous problem, we see that given an odd number a, we
can create a Pythagorean triple:
1 2
1 2
1 2
1 2
a, (a − 1), (a − 1) + 1 = a, (a − 1), (a + 1) .
2
2
2
2
2
We can even do a bit better and actually enumerate these triples by
writing the odd number a as a = 2j + 1, with j = 1, 2, 3, . . . Then, we
have


1
1
2j + 1, ((2j + 1)2 − 1), ((2j + 1)2 + 1) = 2j + 1, 2j 2 + 2j , 2j 2 + 2j + 1
| {z } | {z } |
{z
}
2
2
a
b
c
We check that this always is a Pythagorean triple:
a2 + b2 = (2j + 1)2 + (2j 2 + 2j)2
= 4j 2 + 4j + 1 + 4j 4 + 8j 3 + 4j 2
= 4j 4 + 8j 3 + 8j 2 + 4j + 1
= (2j 2 + 2j + 1)2
= c2
• A natural question at this point is: “Have now found ALL the Pythagorean
triples?” Unfortunately (or fortunately, depending on your point of view), the
following table gives another pattern of Pythagorean triples.
a
b
c
4
3
5
8
15
17
12 35
37
16 63
65
20 99 101
24 143 145
Spend some time determining how these triples are related. Can you determine
the next three triples in this pattern? (Be sure to check that your triples are
Pythagorean triples!)
The next three triples are (28, 195, 197), (32, 255, 257), and (36, 323, 325).
The first thing we notice is that the a values are running through the
positive multiples of 4. We also notice that c = b + 2 and that b and c
are, respectively, 1 more than and 1 less than a perfect square. What
perfect square? Well, c = b + 2 implies that
1
a2 = (b + 2)2 − b2 = 4b + 4 =⇒ b = a2 − 1.
4
Then, for k = 1, 2, 3, . . ., we have a = 4k and so
1
b = (4k)2 − 1 = 4k 2 − 1 = (2k)2 − 1
4
3
and
c = (2k)2 + 1.
• Use the pattern you found in the previous problem to create a formula for these
triples and will hence generate another infinite list of Pythagorean triples.
Prove that your formula always gives a Pythagorean triple.
As we saw in the previous problem, these triples are given by
(|{z}
4k , (2k)2 − 1, (2k)2 + 1).
| {z } | {z }
a
b
c
We check:
a2 + b2 = (4k)2 + ((2k)2 − 1)2
= 16k 2 + 16k 4 − 8k 2 + 1
= 16k 4 + 8k 2 + 1
= ((2k)2 + 1)2
= c2
A General Formula for Pythagorean Triples
We will see how the following formula is derived in the Rational Points on the Unit
Circle section, but for now we will work with this.
(q 2 − p2 , 2pq, p2 + q 2 ) is a Pythagorean triple whenever p and q are positive
integers with q > p. Furthermore, every Pythagorean triple is similar to a
triple of this form.
• Use this formula to find a few Pythagorean triples that you have not yet seen in
this investigation.
Many possible answers, such as
• Prove that, if p, q and k are positive integers with q > p, then
(k(q 2 − p2 ), 2kpq, k(p2 + q 2 )) is a Pythagorean triple.
(k(q 2 − p2 ))2 + (2kpq)2 = k 2 ((q 2 − p2 )2 + 4k 2 p2 q 2
= k 2 (q 4 − 2p2 q 2 + p4 + 4p2 q 2 )
= k 2 (q 4 + 2p2 q 2 + p4 )
= k 2 (q 2 + p2 )2
= (k(p2 + q 2 ))2
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Rational Points on the Unit Circle
In this section, we investigate rational points on the unit circle and make the link to
Pythagorean triples. A point (x, y), on the unit circle is rational if x and y are
rational numbers satisfying x2 + y 2 = 1. Similarly, (x, y) is an integer point on the
unit circle if x and y are integers with x2 + y 2 = 1.
• Find all integer points on the unit circle. How do you know that these are all of
them?
The only integer points on the unit circle are (−1, 0), (0, −1), (1, 0) and
(0, 1). These are the only integer points there are only four integer
points within distance 1 of the origin.
• Find some rational points on the unit circle. For instance, can you have x = 1/2
for a rational point on the unit circle? Explain why or why not.
p
√
If x = 1/2, then y = ± 1 − (1/2)2 = ± 3/2 is not rational. Some
rational points include (3/5, 4/5), (5/13, 12/13), and (8/17, 15/17). Do
you notice anything familiar about these points?
• Now, consider the integer point (−1, 0) on the unit circle. Draw a line through
(−1, 0) and any other point (x, y) on the unit circle. Go to
http://faculty.ithaca.edu/dabrown/docs/pythagorean/rational/
and download the file shown. Double-click to decompress and then load the file
in a browser window.
• Drag the slope slider to change the slope of the this line. If the slope of this line
is m, then write down the equation of this line.
The line between (−1, 0) and (x, y), with slope m has equation
y = m(x + 1).
• Show that if (x, y) is a rational point on the unit circle, the slope of the line
above must also be rational if x 6= −1.
The slope, m, of this line is
m=
y
y−0
=
.
x − (−1)
x+1
So, as long as x 6= −1, this is the quotient of two rational numbers
(with non-zero denominator) and hence is rational.
• Set the slope equal to 0.5 and observe the x and y coordinates of the
intersection point. If you plug these coordinates into the equation of the unit
circle and “clear the denominators”, what do you get? Click the the box show
triangle for a clue. Try this with a few other points.
5
When m = .5, the point of intersection with the unit circle is (3/5, 4/5).
Plugging into the equation of the circle and clearing the denominator
yields
2 2
4
3
+
= 1 =⇒ 32 + 42 = 52
5
5
which is a Pythagorean triple!
• Show that if (a, b, c) is a Pythagorean triple, then the point (a/c, b/c) is a
rational point on the unit circle.
a 2
a2 b 2
a + b = c =⇒ 2 + 2 = 1 =⇒
+
c
c
c
2
2
2
2
b
= 1.
c
Since a, b, and c are integers, a/c and b/c are rational, and (a/c, b/c)
lies on the unit circle.
• Now, show that if (x, y) is the point of intersection of the above line with the
unit circle and the slope m is rational, then x and y are also rational. You can
do this concretely by showing that
x=
1 − m2
1 + m2
and y =
2m
.
1 + m2
We intersect y = m(x + 1) with x2 + y 2 = 1 in the first quadrant.
x2 + y 2 = 1 =⇒ x2 + (m(x + 1))2 = 1
=⇒ x2 + m2 x2 + 2m2 x + m2 − 1 = 0
=⇒ (1 + m2 )x2 + (2m2 )x + (m2 − 1) = 0
p
−2m2 ± 4m4 − 4(1 + m2 )(m2 − 1)
=⇒ x =
2(1 + m2 )
p
−2m2 ± 4m4 − 4(m4 − 1)
=⇒ x =
2(1 + m2 )
√
−2m2 ± 4
=⇒ x =
2(1 + m2 )
−2m2 ± 2
=⇒ x =
2(1 + m2 )
−m2 − 1 1 − m2
=⇒ x =
,
1 + m2 1 + m2
1 − m2
=⇒ x = −1,
1 + m2
6
1 − m2
and
1 + m2
2m
1 − m2 1 + m2
1 − m2
=
+1 =m
+
.
y = m(x + 1) = m
2
2
2
1+m
1+m
1+m
1 + m2
Working in the first quadrant, x =
• Recall that a rational number m can be expressed as p/q for integers p and
q 6= 0. Use this fact and the previous problem to express rational points (x, y) on
the unit circle in terms of p and q. Reduce the fractions as much as possible and
recover the Pythagorean Triples Formula from earlier.
p
Setting m = ,
q
1−m
x=
=
1 + m2
1−
2m
=
y=
1 + m2
2
2
1+
and
Thus,
q 2 − p2
p2 + q 2
2
+
2pq
2
p + q2
2
2
p
q
2 =
p
q
q 2 − p2
p2 + q 2
1+
p
q
2 =
p
q
2pq
.
+ q2
p2
= 1 =⇒ (q 2 − p2 )2 + (2pq)2 = (p2 + q 2 )2 .
So, (q 2 − p2 , 2pq, p2 + q 2 ) gives the general formula for a Pythagorean
triple. And, the geometric construction shows that there is a one-to-one
correspondence between the Pythagorean triples and the rational
points on the unit circle (in the first quadrant).
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