The Right- Angle Triangle and the Pythagorean Triples
INTRODUCTION:
It can be said that the Pythagorean Triple was derived by the Greek Philosopher and
Mathematician Pythagoras and it is closely associated to the right-angled triangle. The
Pythagorean triple represents three positive integers namely 𝑎, 𝑏 and 𝑐 where 𝑎2 + 𝑏 2 = 𝑐 2 and
they can be written as (𝑎, 𝑏, 𝑐)
It can be said that the Pythagorean Triples were derived from Pythagoras’ theorem, a very simple
theorem which is widely used when right-angled triangles are concerned and it states that the
sum of the squares of the sides of a right-angled triangle is equal to the square of the hypotenuse
(longest side). Thus, the formula 𝑎2 = 𝑏 2 + 𝑐 2 where a is the hypotenuse. As it can be seen this
formula is quite similar to that of the Pythagorean Triples.
AIM:
The aim of this investigation is to find out general formula(s) that will help create an infinite
number of Pythagorean Triples aside the primitive Pythagorean Triples.
RATIONALE:
The formula of Pythagoras’ Theorem is the simplest theorems to understand and is widely used
even at the secondary level of education. I had no idea of the existence of the Pythagorean Triple
until we treated Trigonometry in the International Baccalaureate Mathematics class. I read about
the Pythagorean Triples in detail upon research due to my curiosity when the teacher mentioned
it in class when solving for the hypotenuse of a right-angled triangle. I was gravely intrigued by
it. I tried to derive other Pythagorean triples after gaining knowledge on what the primitive
Pythagorean Triples were but this was to no avail thus I realized that the triples do not work on
any random number and it is with this struggle that I chose this topic for my investigation and
hence to find a formula that helps to create an infinite number of Pythagorean triples; and how it
can be used outside the classroom.
HOW DID THE PYTHAGOREAN TRIPLES COME TO BE?
Figure 1:
𝑎
𝑏
Where a, b and c are the lengths of the sides of the squares.
The Pythagorean Triples will represent three positive integers that will satisfy the formula
𝑎2 + 𝑏 2 = 𝑐 2 thus, the square of the hypotenuse (longest side) 𝒄 will be equal to the square of
the sides 𝒂 and 𝒃.
9
3
+
16
=
25
4
5
Therefore:
32 + 42 = 52
9 + 16 = 25
Thus (3, 4, 5) are Pythagorean triples.
DERIVATION OF PYTHAGOREAN TRIPLES
Generally, when trying to derive the Pythagorean triple, we begin by finding the algebraic sum
of the squares of the two smaller numbers which in turn will be used to derive the square of the
hypotenuse that satisfy the condition of the square of the two smaller numbers.
The square of one of the Pythagorean Triples can either be an even or an odd number. Thus
supposing 𝑎2 is the square of one of the Pythagorean Triples and 𝑏 is the second triple.
Therefore:
Using a case where 𝑎2 is odd:
𝑎2 = 2𝑏 + 1 When it is odd,
Making b the subject:
𝑎2 − 1 = 2𝑏
Divide both sides by 2:
(𝑎2 − 1) 2𝑏
=
2
2
∴𝑏=
(𝑎2 − 1)
2
Making 𝑏 + 1 the subject:
Dividing both sides by 2:
𝑎2 2𝑏 1
=
+
2
2 2
1
Adding 2 to both sides
𝑎2 1
1 1
+ =𝑏+ +
2 2
2 2
(𝑎2 + 1)
𝑏+1=
2
The hypotenuse should be:
𝑎2 + 𝑏 2 = 𝑏 2 + 2𝑏 + 1
𝑎2 + 𝑏 2 = (𝑏 + 1)2
The formula 𝑎2 = 2𝑏 + 1 suggests that 2𝑛 + 1 has to be a square number. Therefore, I
substituted a few odd square numbers into the formulas above and the results are shown in the
table below. Whiles trying this formula out, I recognized a pattern that once 𝑎2 was odd; the
value of 𝑏 was always even.
For example: When
𝑎2 = 9
(9 − 1)
𝑏=
=4
2
𝑏+1=4+1
𝑏=4
Using the odd square numbers between 1 − 20:
Figure 2:
𝒂𝟐
𝒂
𝒃
𝒃𝟐
(𝒃 + 𝟏)
(𝒃 + 𝟏)𝟐
𝒂𝟐 + 𝒃𝟐 = (𝒃 + 𝟏)𝟐
9
3
4
16
5
25
9 + 16 = 25
25
5
12
144
13
169
25 + 144 = 169
49
7
24
576
25
625
49 + 576 = 625
81
9
40
1600
41
1681
81 + 1600 = 1681
121
11
60
3600
61
3721
121 + 3600 = 3721
169
13
84
7056
85
7225
169 + 7056 = 7225
225
15
112
12544
113
12769
225 + 12544 = 12769
289
17
144
20736
145
21025
289 + 20736 = 21025
361
19
180
32400
181
32761
361 + 32400 = 32761
I have been able to derive a formula for odd numbers; however, it is not certain that this formula
works for all odd numbers. I realized a unique pattern about the table above; for all the triples
obtained, when arranged in order of magnitude, have a difference of 1 between the last two
numbers. Thus, a triple like 9, 12, 15 will not be obtained even though 9 is an odd square
number. This pattern was unique but was the limitation of my formula.
I have been able to derive a general formula which can help generate a number of triples;
however, the formula can only be applied if at least one of the triples is known, like 𝑎2 and this
known triple must specifically be an odd square number. This makes my formula somewhat
difficult to one that does not have prior knowledge to what a square odd number is. Therefore, to
eliminate this limitation, I chose to generate another formula that can help derive triples without
prior knowledge of one of the triples, meaning a formula that enables the generation of triples
from any real number.
From my first formula, the triple 𝑏 is restricted to even numbers, so
𝑏 = 2𝑥, where 𝑥 is a real number,
𝑎2 = 2𝑏 + 1, then,
𝑎2 = 2(2𝑥) + 1
= 4𝑥 + 1
When 1 is substituted into the equation for 𝑎2 , an odd square number is not obtained, implying
that 𝑏 has a domain which will depend on the domain for 𝑥. Hence, I decided to find the domain
for 𝑥.
Domain for 𝒙
Figure 3:
𝒙
𝒂𝟐
2
9
6
25
12
49
20
81
30
121
42
169
56
225
72
289
90
361
Figure 4:
2
6
12
20
30
42
Sequence of 𝒙
4
First difference
6
8
2
Second Difference
10
2
2
12
2
It can be seen from the table above that the first difference of the 𝑥 values are 4, 6, 8, 10, and
12 respectively while the second difference is 2 all through. This shows that the domain for 𝑏 is
in the form of a quadratic function thus 𝑥 = 𝑎𝑛2 + 𝑏𝑛 + 𝑐. The 𝑥 values can be generated in
terms of 𝑎, 𝑏 and 𝑐 as shown in the tables below.
Figure 5:
𝒏
1
2
3
4
5
6
9𝑎 + 3𝑏 + 𝑐
16𝑎 + 4𝑏 + 𝑐
25𝑎 + 5𝑏 + 𝑐
36𝑎 + 6𝑏 + 𝑐
Sequence
𝟐
𝑎 + 𝑏 + 𝑐 4𝑎 + 2𝑏 + 𝑐
𝒂𝒏 + 𝒃𝒏 + 𝒄
3𝑎 + 𝑏
First difference
Second Difference
5𝑎 + 𝑏
2𝑎
7𝑎 + 𝑏
2𝑎
9𝑎 + 𝑏
2𝑎
11𝑎 + 𝑏
2𝑎
2𝑎 = The second difference of the sequence, then,
2𝑎 = 2
𝑎=1
3𝑎 + 𝑏 = The first value of the first difference, then,
3𝑎 + 𝑏 = 4
3(1) + 𝑏 = 4
𝑏 =4−3=1
𝑎 + 𝑏 + 𝑐 = The first term of the sequence, then,
1+1+𝑐 =2
𝑐 = 2−2 =0
Substituting the values a, b, c into the quadratic formula 𝑎𝑛2 + 𝑏𝑛 + 𝑐,the domain of
𝑥 = 𝑛2 + 𝑛, where 𝑏 is the 𝑛𝑡ℎ term of the sequence.
Hence,
𝑏 = 2𝑥
𝑏 = 2(𝑛2 + 𝑛)
𝑏 2 = [2(𝑛2 + 𝑛)]2
Since 𝑏 + 1 = 2(𝑛2 + 𝑛) + 1
⟹ (𝑏 + 1)2 = [2(𝑛2 + 𝑛) + 1]2
𝑎2 = 4𝑥 + 1
𝑎2 = 4(𝑛2 + 𝑛) + 1
The first general formula 𝑎2 + 𝑏 2 = (𝑏 + 1)2 , will now become
[4(𝑛2 + 𝑛) + 1] + [2(𝑛2 + 𝑛)]2 = [2(𝑛2 + 𝑛) + 1]2
I will now illustrate this formula on a right-angled triangle to show how the Pythagorean Triples
can be obtained
Figure 6:
𝑏 + 1 = 2(𝑛2 + 𝑛) + 1
𝑚 = √(4(𝑛2 + 𝑛) + 1)
𝑏 = 2(𝑛2 + 𝑛)
I will also use Microsoft excel to generate Pythagorean triples using the real numbers from 1 −
20 in order to prove the certainty of this formula working for all real numbers.
Figure 7:
Usage of the formulas illustrated on the right-angled triangle above.
Figure 8:
The triple generated by using the formulas displayed on the right-angled triangle above has
been put in the table with black, visible borders.
Thus far, the formula seems to work. However this was tested using only the first 20 real
numbers so in order to test the exactitude of my formula and validate its ability to work on all
real numbers, I decided to use mathematical induction, a technique for proving results for natural
numbers, which I learnt about after staying in a Mathematics Higher Level class during my free
time. I studied Mathematical induction hence which I will utilize this topic as a method of
proving whether or not the formula:
2
[4(𝑛 + 𝑛) + 1] + [2(𝑛2 + 𝑛)]2 = [2(𝑛2 + 𝑛) + 1]2 works for all real numbers.
USING MATHEMATICAL INDUCTION TO PROVE THE FORMULA:
[𝟒(𝒏𝟐 + 𝒏) + 𝟏] + [𝟐(𝒏𝟐 + 𝒏)]𝟐 = [𝟐(𝒏𝟐 + 𝒏) + 𝟏]𝟐 IS TRUE FOR ALL REAL
NUMBERS
Let 𝑊(𝑛): 4(𝑛2 + 𝑛) + 1 + [2(𝑛2 + 𝑛)]2 = [2(𝑛2 + 𝑛) + 1]2
When 𝑛 = 1,
𝑤(1):
𝐿𝐻𝑆 = 4(12 + 1) + 1 + [2(12 + 1)]2 = 25
𝑅𝐻𝑆 = [2(12 + 1) + 1]2 = 25
∴ 𝐿𝐻𝑆 = 𝑅𝐻𝑆
This means that for the formula, 𝑊(𝑛) = 4(𝑛2 + 𝑛) + 1 + [2(𝑛2 + 𝑛)]2 = [2(𝑛2 + 𝑛) + 1]2
it is true that for 𝑛 = 1 ,𝑤(1) is true. Therefore assuming 𝑛 = 𝑘, where 𝑤(𝑘) is also true, then
4(𝑘 2 + 𝑘) + 1 + [2(𝑘 2 + 𝑘)]2 = [2(𝑘 2 + 𝑘) + 1]2
∴ 𝑊ℎ𝑒𝑛 𝑛 = 𝑘 + 1
𝐿𝐻𝑆 = 4[(𝑘 + 1) + (𝑘 + 1)] + 1 + [2((𝑘 + 1)2 + (𝑘 + 1))]2
2
= [2((𝑘 + 1)2 + (𝑘 + 1))]2 + 4[(𝑘 + 1)2 + (𝑘 + 1)] + 1
= [2((𝑘 + 1)2 + (𝑘 + 1))]2 + 2[2((𝑘 + 1)2 + (𝑘 + 1))] + 1
Let = [2((𝑘 + 1)2 + (𝑘 + 1))] , then
𝐿𝐻𝑆 = 𝑑 2 + 2𝑑 + 1
= (𝑑 + 1)2
= [2((𝑘 + 1)2 + (𝑘 + 1)) + 1]2
𝑅𝐻𝑆 = [2((𝑘 + 1)2 + (𝑘 + 1) + 1]2
∴ 𝐿𝐻𝑆 = 𝑅𝐻𝑆
Since the 𝐿𝐻𝑆 = 𝑅𝐻𝑆 in both 𝑛 = 1 and 𝑛 = 𝑘 + 1; 𝑤(1) is true and also true for 𝑤(𝑘 + 1)
therefore in any instance where it is true for w(k), then the formula 𝑊(𝑛) = 4(𝑛2 + 𝑛) + 1 +
[2(𝑛2 + 𝑛)]2 = [2(𝑛2 + 𝑛) + 1]2 is true for all real numbers as well. With the use of
mathematical induction, it has been proven that this formula indeed does work for all real
numbers.
EUCLID’S FORMULA
Euclid's formula for a Pythagorean triple states that: 𝑎 = 2𝑚𝑛, 𝑏 = 𝑚2 − 𝑛2 , 𝑐 = 𝑚2 + 𝑛2 . This
formula allows us to use two unidentified constants 𝑚 and n to generate a Pythagorean triple. In
this case, 𝑚2 > 𝑛2 therefore 𝑚 > 𝑛.
DERIVING EUCLID’S FORMULA
Assuming the right-angled triangle has the sides 𝑎, 𝑏 and 𝑐:
Figure 9:
𝑐
𝑎
𝑏
From the theory of the Pythagoras theorem, I can conclude that 𝑐 2 = 𝑎2 + 𝑏 2 . Making 𝑎 2 the
subject gives 𝑎2 = 𝑐 2 − 𝑏 2 .
Assuming 𝑎 is even:
𝑎2 = (𝑐 − 𝑏)(𝑐 + 𝑏)
𝑎
(𝑐 + 𝑏)
=
(𝑐 − 𝑏)
𝑎
The right hand side is the rational part of this equation therefore it can be equated to
𝑚 and 𝑛 are in their lowest terms.
𝑚
𝑛
where
(𝑐 + 𝑏) 𝑚
= → 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1
𝑎
𝑛
Since
(𝑐+𝑏)
𝑎
=
𝑚
𝑛
𝑎
then (𝑐−𝑏) =
𝑚
𝑛
In order to make the left hand side rational, I flipped the equation therefore,
(𝑐 − 𝑏) 𝑛
= → 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2
𝑎
𝑚
Adding Equation 1 and Equation 2:
𝑐 𝑐
𝑏 𝑏
𝑚 𝑛
( + )+( − )= +
𝑎 𝑎
𝑎 𝑎
𝑛 𝑚
=
2𝑐 𝑚2 + 𝑛2
=
𝑎
𝑚𝑛
=
𝑐 𝑚 2 + 𝑛2
=
𝑎
2𝑚𝑛
Subtracting Equation 1and Equation 2:
𝑐 𝑐
𝑏 𝑏
𝑚 𝑚
( − )+( + )=( − )
𝑎 𝑎
𝑎 𝑎
𝑛 𝑛
2𝑏 𝑚2 − 𝑛2
=
=
𝑎
𝑥𝑦
2
𝑏 𝑚 − 𝑛2
= =
𝑎
2𝑥𝑦
Assuming that the left hand side of the numerator is equal to its corresponding right hand side
numerator and the left hand side denominator is also equal to its corresponding right hand side
denominator, then this gives 𝑎 = 2𝑥𝑦, 𝑏 = 𝑥 2 − 𝑦 2 , 𝑐 = 𝑥 2 + 𝑦 2 . This formula is known as the
Euclid formula.
USING EUCLID’S FORMULA TO GENERATE A TRIPLE
Using 𝑎 = 2𝑥𝑦, I can attain different factors and use these factors to find triples.
Supposing 20 was a number
𝑎 = 20 then 𝑥𝑦 = 10
Possible factors of 10 are: (2,5) and (10,1)
Using the equation 𝑏 = 𝑥 2 − 𝑦 2 and 𝑐 = 𝑥 2 + 𝑦 2, I will be able to derive the other two triples.
𝑏 = 52 − 22 = 25 − 4 = 21
𝑐 = 52 + 22 = 25 + 4 = 29
And
𝑏 = 102 − 12 = 99
𝑐 = 102 + 12 = 101
In order to make sure that these numbers are triples using the Pythagorean theorem 𝑎2 + 𝑏 2 =
𝑐2:
202 + 992 = 10201
√10201 = 101
When I substituted the numbers above into the Pythagoras theorem, I saw that truly
𝑐 2 = 𝑎2 + 𝑏 2 . This confirmed that the numbers obtained were triples.
CONCLUSION:
On the basis of odd and even numbers, I have been able to derive a general formula that can
generate an infinite amount of Pythagorean triples without the need to have prior knowledge of
one of the triples. Assuming 𝑎2 is an odd square number and 𝑏 is an even number from the
Pythagorean triples equation 𝑎2 = 𝑏 2 + 𝑐 2 I have been able to derive the general
formula[4(𝑛2 + 𝑛) + 1] + [2(𝑛2 + 𝑛)]2 = [2(𝑛2 + 𝑛) + 1]2 for all real numbers. This formula
has been proven valid with the help of mathematical induction. Euclid’s formula also proved that
indeed there is a way to derive Pythagorean triples using two unknown constants. It is the safe to
say that the aim of this investigation has been achieved.
REFLECTION:
The aim of this investigation was to find out general formula(s) that will help create an infinite
number of Pythagorean Triples aside the primitive Pythagorean Triples and its significance will
help people who are math inclined, especially students, to be able to create Pythagorean Triples
using any real number. It will save them from having to memorize countless triples when
needed. This work can be improved by investigating the different applications of the
Pythagorean triples, with the use if the general formula obtained in order to further solidify the
value of the formula. The general formula [4(𝑛2 + 𝑛) + 1] + [2(𝑛2 + 𝑛)]2 = [2(𝑛2 + 𝑛) + 1]2
attained, at first glance, seems very lengthy thus one could invent a shorter and simpler general
formula hence minimizing the difficulty in the memorization of such a lengthy formula.
References
Lawrence O. Cannon. “Pythagorean Triples: Avenues for Exploration.” The Mathematics
Teacher, vol. 105, no. 4, 2011, pp. 311–315. JSTOR, JSTOR . n.d. 26 Sep. 2017.
<www.jstor.org/stable/10.5951/mathteacher.105.4.0311>.
Tobin-Campbell, C. Systems of Pythagorean triples. n.d. 26 Sep. 2017.
<https://www.whitman.edu/Documents/Academics/Mathematics/SeniorProject_ChrisTobinCampbell.pdf>
“Pythagorean Triples” Teaching Math History. n.d. 26 Sep. 2017.
<http://www.wwu.edu/teachingmathhistory/docs/psfile/Pythag3-teacher.pdf >
“Pythagorean Triples” Teaching Math History. n.d. 27 Sep. 2017.
<http://www.wwu.edu/teachingmathhistory/docs/psfile/Pythag3-student.pdf -9/4/2017>
“Mathematical Induction” Tutorials Point. n.d. 27 Sep. 2017.
<https://www.tutorialspoint.com/discrete_mathematics/discrete_mathematical_induction.htm>
MIKE MOLONY. “Generating Pythagorean Triples” Dreamshire. n.d. 14 Nov. 2017.
<https://blog.dreamshire.com/generating-pythagorean-triples/>