New Pythagorean Triples copy

Pythagorean Triples
Drew Erskine & Nicole Dardano
November 28, 2015
1 Preface
A man named Pythagoras was born in 596 B.C.E on the islands of Samos, Greece. He was the
founder of a group called ‘The Brotherhood of Pythagoreans’ where they studied mathematics and
spread their knowledge of math to the people around them. Though much of Pythagoras’ work was
kept secret in the group, he was known to have proved the Pythagorean Theorem. This was one
of the most influential findings of Pythagoras. Many of us can recognize this theorem because at
some point in our education it was taught to us. The Pythagorean Theorem states that if you have
a right triangle with legs a, b and hypotenuse c (where a, b, c ∈ N) then the triple (a, b, c) satisfies:
a2
+ b2
= c2
Only we arent really interested in just looking at right triangles, instead we intend to observe
Pythagorean triples from triangles where the hypotenuse comes from a 60◦ angle and then later on
a θ◦ angle.
Pythagorean triples are when — in the equation above — a, b, and c are whole numbers. There
are a few interesting properties of triples that we should keep in mind. For instance, there are two
types of triples. There are those who belong in a family of triples, then there are your family of
basic triples. Take this table of triples for example:
32 + 42 = 52 62 + 82 = 102 . . . (3n)2 + (4n)2 = (5n)2
52 + 122 = 132 102 + 242 = 262 . . . (5n)2 + (12n)2 = (13n)2
72 + 242 = 252 142 + 482 = 502 . . . (7n)2 + (24n)2 = (25n)2
.
.
.
.
.
.
.
.
.
.
.
.
a2 + b2 = c2 . . . . . . (an)2 + (bn)2 = (cn)2
This table demonstrates a lot of key questions and components to Pythagorean triples. Lets break
down this table and explain its structures. All triples to the left of the divider, are not generated
by multiplying through by a constant n and a, b, c have a GCD = 1. We call these families of triples
“Basic Triples”. Every triple to the right of the divider are not basic triples, this is because we
have manipulated the numbers by multiplying through by a constant n. These families of triples
come from a basic triple. For simplicity lets call these “Generated Triples”. One might notice that
through the generated triples, there are an infinite amount of triples. But what can we say about
our family of basic triples?
1

Throughout our research we have created more information about basic triples and their properties.
Along with findings about basic triples, we began to understand the geometric structure of a triple
and how they are generated algebraically and geometrically. With the use of basic algebra and
some geometry, we are telling our story of Pythagorean triples.
2 The Equivalence Relation
We will first start by providing more information about the families of triples before we get involved
more in how we generate triples.
Theorem 2.1. We can say two Pythagorean triples, (a, b, c) ∼ (a , b , c ), if and only if a = ka ,
b = kb , and c = kc , for some k ∈ Q+.
Proof. To prove an equivalence relation, we must prove that reflexivity, symmetry and transitivity
all hold.
1. Reflexivity (a, b, c, ) ∼ (a, b, c) a = a, b = b, and c = c, therefore (a, b, c, ) ∼ (a, b, c) and
reflexivity holds.
2. Symmetry If (a, b, c) ∼ (a , b , c ) then (a , b , c ) ∼ (a, b, c). By definition of this relation,
a = ka , b = kb , and c = kc . And reflexively, ka = a, kb = b, and kc = c. Therefore, a = 1
k a,
b = 1
k b, and c = 1
k c, and so symmetry holds.
3. Transitivity If (a, b, c) ∼ (a , b , c ) and (a , b , c ) ∼ (a , b , c ), then (a, b, c) ∼ (a , b , c ). By
definition, a = ka , b = kb , and c = kc . Given an m ∈ Q+, a = ma , b = mb , and c = mc . By
substitution, a = kma , b = kmb and c = kmc . Let km = n then a = na , b = nb and c = nc ,
therefore by definition, (a, b, c) ∼ (a , b , c ) and transitivity holds.
3 Rational Triples
As a precursor of our investigation, we did not start with whole numbers but instead we investigated
rational numbers. When looking at:
a2
+ b2
= c2
when you divide through by c2 you get:
a2
c2
+
b2
c2
= 1
By letting a
c = x and b
c = y, then in terms of algebraic geometry, x2 + y2 = 1 is the equation of the
unit circle. So by looking at each rational point on the unit circle [i.e. ((a
c ), (b
c ))], we are guaranteed
to find a triple. We know we can get all rational points because of the following:
2

(a) Line from (-1,0) to (x, y)
The reason why we consider starting the integer point (-1,0) and going to a rational (x, y) point
in the first quadrant on the unit circle is really subtle. As promised, the idea of choosing a point
(x, y) in the first quadrant is because every point is positive. Also there are the same amount of
rational points in the first quadrant then there are in the second. This is true because the second
quadrant is a reflection of the first across the y-axis. Using the slope of this line, the equations can
be compared. But the real reason why we choose the first quadrant is because, when you draw a
line from (−1, 0) to the point (x, y), you generate a line with a positive slope. This is crucial when
it comes to the algebra because now we do not have to worry about negative signs floating around.
During our investigation, we realized there must be some type of relationship with rational triples
(x2 + y2 = 1) and Pythagorean triples (a2 + b2 = c2) since we have already created a link between
their equations.
Theorem 3.1. Every basic triple yields a rational point on the unit circle and every rational point
on the unit circle yields a triple.
Proof. If we have any basic triple, (a, b, c), where a2 + b2 = c2, to find a rational point that
corresponds to it, you can convert a2 + b2 = c2 to x2 + y2 = 1, by simply dividing each term by c2.
So essentially you have a2
c2 + b2
c2 = 12 which can be written as (a
c )2 + (b
c )2 = 1, therefore yielding a
rational point, (a
c , b
c) satisfying the equation x2 + y2 = 1, so it is on the unit circle.
Describing the relationship of rational point to Pythagorean triples requires a little more algebra to
be convinced. Let x = r1 and y = r2 be rational point on the unit circle that satisfy the equation,
x2 + y2 = 1. Giving us (r1)2 + (r2)2 = 1. Since r1 and r2 are rational numbers, we can write:
r1 = m
n and r2 = p
q where m, n, p, q ∈ Z+, n, q = 0, m < n and p < q. So,
1 = (
m
n
)2
+ (
p
q
)2
= (
mq
nq
)2
+ (
np
nq
)2
=
m2q2
n2q2
+
n2p2
n2q2
n2
q2
= m2
q2
+ n2
p2
Which can also be written (mq)2 + (np)2 = (nq)2. With this form, we can see that mq = a, np = b
3

and nq = c. Thus, the relationship from rational points to Pythagorean triples is:
(
m
n
,
p
q
) → (mq, np, nq)
and this conclusion proves every rational point guarantees a triple. This triple is not necessarily
a basic triple, it may be a scaled triple belonging to some family. In order for (mq, np, nq) to be
basic, GCD(mq, np, nq) = 1.
3.1 Finding Rational Points
To make things more simplistic, as we previously stated, we can just look at all rational points, so
we substitute ((a
c ), (b
c)) with the point (x, y) on the unit circle that is in the first quadrant. What
it takes to generate a triangle is you find a point on the curve that is rational and generate the
line from (-1,0) to said point. Our objective now is to find every rational sloped line getting us a
rational (x, y) given:
x2
+ y2
= 1
Lets look at some examples, for a slope of m = 1
2, the corresponding (x, y) is (3
5, 4
5), and yields an
equation of y = 1
2x + 1
2.
(b) (3
5 , 4
5 ) with y = 1
2 x + 1
2
When m = 1, (x, y) is (0,1) then y = x + 1. And when m =3
2, (x, y) is (−5
13 , 12
13) which can be found
in the first quadrant as ( 5
13, 12
13) yeilding the slope m =2
3 then y = 2
3x + 2
3.
Proposition 3.2. The slope of the line from (−1, 0) to (x, y) is rational, if and only if, x and y
are rational.
Proof. Given this, we can say y = m(x + 1) or y = mx + m. So if we consider the equation
x2 + y2 = 1, we can substitute for y and find x remembering that m > 0.
4

1 = x2
+ y2
= x2
+ (m(x + 1))2
= x2
+ (mx + m)2
= x2
+ m2
x2
+ 2m2
x + m2
0 = (1 + m2
)x2
+ (2m2
)x + (m2
− 1)
By using the quadratic formula,
x =
−2m2 ± 4m4 − 4(1 + m2)(m2 − 1)
2(1 + m2)
=
−2m2 ± 4m4 − 4(m4 − 1)
2(1 + m2)
=
−2m2 ±
√
4
2(1 + m2)
=
−2m2 ± 2
2(1 + m2)
=
−m2 ± 1
1 + m2
So then −(m2+1)
1+m2 and 1−m2
1+m2 . This proves that if m is rational, then x will be rational and if x is
rational, then m is rational, because rational numbers are closed under multiplication and division.
If we substitute −(m2+1)
1+m2 into y = m(x + 1) then y = 0. If x = 1−m2
1+m2 we get:
y = m(
1 − m2
1 + m2
+ 1)
= m(
1 − m2
1 + m2
+
1 + m2
1 + m2
)
= m(
1 − m2 + 1 + m2
1 + m2
)
= m(
2
1 + m2
)
= m(
2m
1 + m2
)
=
2m
1 + m2
5

This proves that if we have a point (x, y) as the point of intersection with the line starting at (−1, 0)
and a rational slope m, then y will also be rational. So if we include our previous results, if the
slope is rational, both x and y will be rational with x = 1−m2
1+m2 and y = 2m
1+m2 .
3.2 Finding Formulas
By expressing m as a rational number, m = p
q , where p, q ∈ Z and q = 0 and p < q (because m < 1)
and by using our previous formulas of x and y, we get:
x =
1 − (p
q )2
1 + (p
q )2
y =
2(p
q )
1 + (p
q )2
=
1 − p2
q2
1 + p2
q2
=
2p
q
q2+p2
q2
=
q2−p2
q2
q2+p2
q2
=
2pq
q2 + p2
=
q2 − p2
q2 + p2
We can use these results to create a general formula for Pythagorean triples, x2 + y2 = 1.
1 = (
q2 − p2
q2 + p2
)2
+ (
2pq
q2 + p2
)2
=
(q2 − p2)2 + (2pq)2
(q2 + p2)2
(q2
+ p2
)2
= (q2
− p2
)2
+ (2pq)2
We let q2 − p2 = a, 2pq = b then we get q2 + p2 = c. If we know the rational slope of the line,
going through (-1,0) and a point on the unit circle, we can ﬁnd a Pythagorean triple. Therefore
our formulas for ﬁnding triples are:
a = q2 − p2 b = 2pq c = q2 + p2
So to form a triple, you just need a rational slope, 0 < p
q < 1 and use the formulas.
Example 3.3. Let m = 3
5. So,
a = 52 − 32 = 16
b = 2 ∗ 5 ∗ 3 = 30
c = 32 + 52 = 34
Giving us a (16, 30, 34) triple, which belongs to the family of the basic triple (8, 15, 17) scaled by a
factor of 2.
6

4 60◦
Triangles
The equation for triangles with a 60◦ angle opposite the hypotenuse is very similar to the Pythagorean
Theorem mainly because they both originated from the Law of Cosines:
c2
= a2
+ b2
− 2ab cos θ
For a right angle, because we know cos (90) = 0, we obtain, c2 = a2 + b2. If we do the same for a
60◦ angle, we get c2 = a2 + b2 − 2ab cos (60) and cos (60) = 1
2, we get:
c2
= a2
+ b2
− ab
Which is the equation used to find 60◦-triples. These are similar to Pythagorean Triples in the
sense that (a, b, c) is a triple where a, b, c ∈ N and c is the hypotenuse, or in this case, the length
opposite the 60◦ angle. Unlike the Pythagorean Theorem, the 60◦ equation yields an ellipse. When
you divide the equation by c2 and get 1 = x2 + y2 − xy, which is an ellipse centered around the
origin (recall that x and y are the rational numbers (a
c )2 and (b
c)2).
(c) 1 = x2
+ y2
− xy
Similar to the Unit Circle, the ellipse goes through the point (−1, 0) so we can use this point again
to find all rational points on the ellipse in the first quadrant, where all points are positive. So
again, the equation of the line through (−1, 0) and a point on the ellipse is y = mx + m and with
the same process previously done with right angles, we can prove we can obtain all rational points
on the ellipse. When mx + m is substituted for y in 1 = x2 + y2 − xy, using the quadratic formula
we get,
x1 = −1 x2 =
−m2 + 1
m2 − m + 1
By plugging the x2 into y = mx + m, we find:
y =
2m − m2
m2 − m + 1
We know the point (x, y) is rational because m is rational. This means m can be written as p
q ,
where p, q ∈ N and p < q. We can find x and y more specifically by plugging p
q into our (x, y). By
doing this, we get:
7

x =
q2 − p2
p2 − pq + q2
y =
2pq − p2
p2 − pq + q2
representing (x, y) from a rational slope. Substituting this into our formula for an ellipse, x2 −xy +
y2 = 1, we can find formulas for a, b and c.
a = q2
− p2
b = 2pq − p2
c = p2
− pq + q2
Therefore, just by picking a rational slope where p > q and q = 0 we can find a 60◦-triple.
5 θ◦
Triangles
After examining properties of 90◦ and 60◦ triangles, we began to see some patterns that we can
generalize. Since we can always pick a triangle with θ◦ and two sides, we can use the Law of Cosines
again to generate a formula to find triples. The first realization was that if we chose any angle of
θ we have possibly get an irrational c2. For example, if we chose θ = 45◦,
c2
= a2
+ b2
− 2ab cos (45)
= a2
+ b2
− ab
√
2
So if a, b ∈ N, then c2 and c would be irrational and therefore we would not have rational points
within the first quadrant excluding (±1, 0) and (0, ±1) on the ellipse. In order to avoid this issue,
we have to choose a θ so that cos θ ∈ Q. But instead of several tries of guessing, we chose to work
backwards and chose what we wanted our cos θ to equal first.
Since −1 < cos θ < 1 we have −1 < x < 1 to choose from. First, we let cos θ = 2
3. The
arccos (2
3) ≈ 48.1915◦. Using the Law of Cosines, our equation is: c2 = a2 + b2 − 4
3ab and when we
rationalize it, 1 = x2 + y2 − 4
3xy. Here is a picture of the corresponding ellipse with the original in
the background:
If we go through the same process as we did with 90◦ and 60◦ equations we can find any rational
point (x, y). Again, we can use the point (−1, 0) to find a slope through a point on the ellipse.
From this,
x =
1 − m2
m2 − 4
3m + 1
y =
2m − 4
3m2
m2 − 4
3m + 1
8

(d) 1 = x2
+ y2
− 4
3 xy
Using a rational m where m = p
q , we get
x =
(q2 − p2)
3p2 − 4pq + 3q2
y =
2(3pq − 2p2)
3p2 − 4pq + 3q2
And from these equations, we can ﬁnd triples using:
a = q2
− p2
b = 2(3pq − 2p2
)
c = 3p2
− 4pq + 3q2
.
A general case for when cos θ = s
t where s, t ∈ N and |s| < t. Using Law of Cosines
c2
= a2
+ b2
−
2s
t
ab
For all our rational points (x, y),
x =
1 − m2
m2 − 2s
t m + 1
y =
2m − 2s
t m2
m2 − 2s
t m + 1
With a rational slope, p
q , we can also generate any triples with
a = q2
− p2
b = 2pq −
2s
t
p2
c = q2
−
2s
t
pq + p2
9

6 Geometric Proof
Throughout our project there has been a lingering problem. We have shown that algebraically,
we can generate triples for the hypotenuse angle equaling 90o, 60o and θo; the question is, can we
generate triples geometrically? It is easy to see for the 90o triangle.
Theorem 6.1. You can generate a Pythagorean triple through the use of geometry
Proof. If we start by looking at our unit circle, we choose the point A on the unit circle where
A = ((a
c )2, (b
c )2) and a, b, c ∈ N, we can draw a line from the origin to the point A.
From there we can drop the perpendicular of point A to the point B = ((a
c )2, 0) this gives us the
lines (O = (0, 0)) AB = (b
c)2, OB = (a
c )2 and because we know that the unit circle has radius 1 we
know that OA = 1
Now that we have all of the components, we can scale the triangle OAB by c2 this gives us the
triangle and the basic triple at the same time.
This means the triangle triple (a, b, c) was generated geometrically with the use of x2 + y2 = 1.
This brings us to our lingering problem. Is there a way to geometrically generate your triangle
triple (a, b, c) for a 60o triangle? What about a θo triangle?
We will now show you the case for the 60o triangle and within the proof it will be easy to see how
to geometrically generate θo triples.
10

Theorem 6.2. You can geometrically generate 60o triangle triples
Proof. Assume the point A is a point on the curve x2 + y2 − xy = 1 where A = ((a
c )2, (b
c)2) and
a, b, c ∈ N. From A we drop the perpendicular to point B = ((a
c )2, 0).
We then make a circle with radius (b
c )2 have point B as the center than pick a point C on this new
circle such that ∠OBC = 60o where O is the origin.
Thus we have created the triangle OBC. We know this to be true because BC = (b
c)2, OB = (a
c )2
and OC = 1 by the Law of Cosines. You technically do not need the Law of Cosines because we
know that, ((a
c )2, (b
c)2) satisﬁes our parameter x2 + y2 − xy = 1. It is now noticeable that when
scaled by c2, OBC is the triangle we are looking for.
11

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