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# Pythagorean Triples – Explanation & Examples

## What is a Pythagorean triple?

**Pythagorean triple (PT) can be defined as a set of three positive whole numbers that perfectly satisfy the Pythagorean theorem: a ^{2} + b^{2 }= c^{2}.**

This set of numbers are usually the three side lengths of a right triangle. Pythagorean triples are represented as: (a, b, c), where, a = one leg; b = another leg; and c = hypotenuse.

*There are two types of Pythagorean triples:*

- Primitive Pythagorean triples
- Non-primitive Pythagorean triples

### Primitive Pythagorean triples

**A primitive Pythagorean triple is a reduced set of the positive values of a, b, and c with a common factor other than 1**. This type of triple is always composed of one even number and two odd numbers.

**For example**, (3, 4, 5) and (5, 12, 13) are examples of primitive Pythagorean triples because each set has a common factor of 1 and also satisfies the

Pythagorean theorem: a^{2} + b^{2 }= c^{2}.

- (3, 4, 5) → GCF =1

a^{2} + b^{2 }= c^{2}

3^{2} + 4^{2 }= 5^{2}

9 + 16 = 25

25 = 25

- (5, 12, 13) → GCF = 1

a^{2} + b^{2 }= c^{2}

5^{2} + 12^{2 }= 13^{2}

25 + 144 = 169

169 = 169

### Non-primitive Pythagorean triples

**A non-primitive Pythagorean triple, also known as the imperative Pythagorean triple, is a set of positive values of a, b, and c with a common factor greater than 1**. In other words, the three sets of positive values in a non-primitive Pythagorean triple are all even numbers.

**Examples of non-primitive Pythagorean triples include**: (6,8,10), (32,60,68), (16, 30, 34) etc.

- (6,8,10) → GCF of 6, 8 and 10 = 2.

a^{2} + b^{2 }= c^{2}

6^{2} + 8^{2 }= 10^{2}

36 + 64 = 100

- = 100
- (32,60,68) → GCF of 32, 60 and 68 = 4

a^{2} + b^{2 }= c^{2}

32^{2} + 60^{2 }= 68^{2}

1,024 + 3,600 = 4,624

4,624 = 4,624

Other examples of commonly used Pythagorean triples include: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25) , (20, 21, 29) , (12, 35, 37), (9, 40, 41), (28, 45, 53), (11, 60, 61), (16, 63, 65), (33, 56, 65), (48, 55, 73), etc.

### Properties of Pythagorean triples

From the above illustration of different types of Pythagorean triples, we make the following **conclusions about Pythagorean triples**:

- A Pythagorean triple cannot have composed of only odd numbers.
- Similarly, a triple a Pythagorean triple can never contain one odd number and two odd numbers.
- If (a, b, c) is a Pythagorean triple, then either a or b is the short or long leg of the triangle, and c is the hypotenuse.

## Pythagorean Triples Formula

The Pythagorean triples formula can generate both primitive Pythagorean triples and non-primitive Pythagorean triples.

Pythagorean triples formula is given as:

(a, b, c) = [ (m^{2} − n^{2}); (2mn); (m^{2} + n^{2})]

Where m and n are two positive integers and m > n

**NOTE**: If one member of the triple is known, we can obtain the remaining members by using the formula: (a, b, c) = [ (m^{2}-1), (2m), (m^{2}+1)].

*Example 1*

What is the Pythagorean triple of two positive numbers, 1 and 2?

__Solution__

Given the Pythagorean triples formula: (a, b, c) = (m^{2} − n^{2}; 2mn; m^{2} + n^{2}), where; m > n.

So, let m = 2 and n = 1.

Substitute the values of m and n into the formula.

⇒ a = 2^{2} − 1^{2} = 4 − 1 = 3

a =3

⇒ b = 2 × 2 × 1 = 4

b = 4

⇒ c = 2^{2} + 1^{2} = 4 + 1 = 5

c = 5

Apply the Pythagorean theorem to verify that (3,4,5) is indeed a Pythagorean triple

⇒ a^{2} + b^{2 }= c^{2}

⇒ 3^{2} + 4^{2 }= 5^{2}

⇒ 9 + 16 = 25

⇒ 25 = 25.

Yes, it worked! Therefore, (3,4,5) is a Pythagorean triple.

*Example 2*

Generate a Pythagorean triple from two integers 5 and 3.

__Solution__

Since m must be greater than n (m > n), let m= 5 and n = 2.

a = m^{2} − n^{2}

⇒a= (5)^{2 }−(3)^{2} = 25−9

= 16

⇒ b = 2mn = 2 x 5 x 3

= 30

⇒ c = m^{2} + n^{2 }= 3^{2 }+ 5^{2}

= 9 + 25

= 34

Hence, (a, b, c) = (16, 30, 34).

Verify the answer.

⇒ a^{2} + b^{2 }= c^{2}

⇒ 16^{2} + 30^{2 }= 34^{2}

⇒ 256 + 900 = 1,156

1,156 = 1,156 (True)

Therefore, (16, 30, 34) is indeed a Pythagorean triple.

*Example 3*

Check if (17, 59, 65) is a Pythagorean triple.

__Solution__

Let, a = 17, b = 59, c = 65.

Test if, a^{2} + b^{2 }= c^{2}.

a^{2} + b^{2 }⇒ 17^{2} + 59^{2 }

⇒ 289 + 3481 = 3770

c^{2 }= 65^{2}

= 4225

Since 3770 ≠ 4225, then (17, 59, 65) is not a Pythagorean triple.

*Example 4*

Find the possible value of ‘a’ in the following Pythagorean triple:(a, 35, 37).

__Solution__

Apply the Pythagorean equation a^{2} + b^{2 }= c^{2}.

a^{2} + 35^{2 }= 37^{2}.

a^{2 }= 37^{2}−35^{2}=144.

√a^{2 }= √144

a = 12.

*Example 5*

Find the Pythagorean triple of a right triangle whose hypotenuse is 17 cm.

__Solution__

(a, b, c) = [ (m^{2}-1), (2m), (m^{2}+1)]

c = 17 = m^{2}+1

17 – 1 = m^{2}

m^{2} = 16

m = 4.

Therefore,

b = 2m = 2 x 4

= 8

a = m^{2} – 1

= 4^{2} – 1

= 15

*Example 6*

The smallest side of a right triangle is 20mm. Find the Pythagorean triple of the triangle.

__Solution__

(a, b, c) =[(2m), (m^{2}-1), (m^{2}+1)]

20 =a = 2m

2m = 20

m =10

Substitute m = 10 into the equation.

b = m^{2} – 1

= 10^{2} – 1= 100 – 1

b = 99

c = m^{2}+1

= 10^{2} + 1

= 100 + 1 = 101

PT = (20, 99, 101)

*Example 7*

Generate a Pythagorean triple from two integers 3 and 10.

__Solution__

(a, b, c) = (m^{2} − n^{2}; 2mn; m^{2} + n^{2}).

a = m^{2} − n^{2}

= 10^{2} – 3^{2} = 100 – 9

= 91.

b = 2mn = 2 x 10 x 3

= 60

c = m^{2} + n^{2}

= 10^{2 }+ 3^{2} = 100 + 9

= 109.

PT = (91, 60,109)

Verify the answer.

a^{2} + b^{2 }= c^{2}.

91^{2} + 60^{2 }= 109^{2}.

8,281+ 3,600=11,881

11,881=11,881 (True)

*Example 8*

Check whether the set (24, 7, 25) is a Pythagorean triple**.**

__Solution__

Let a = 24, b = 7 and c = 25.

By Pythagorean theorem: a^{2} + b^{2} = c^{2}

7^{2} + 24^{2} = 625

49 + 576 = 625 (True)

Therefore, (24, 7, 25) is a Pythagorean triple.

*Example 9*

Find the Pythagorean triplet of a right triangle whose one side is 18 yards.

__Solution__

Given the formula: (a, b, c) = [ (m^{2}-1), (2m), (m^{2}+1)].

Let a or b = 18 yards.

2m = 18

m = 9.

Substitute m = 9 into the formula.

c = m^{2} + 1

= 9^{2 }+ 1 = 81

b or a = m^{2} -1 = 9^{2} -1

= 80

Therefore, the possible triplets are; (80, 18, 81) or (18, 80, 81).