# Generating Pythagorean Triples using a Formula

You can generate a Pythagorean Triple using a formula. The proof of why the following formula works all the time is beyond the scope of this lesson. For our purposes, let us call this the “Pythagorean Triple Formula”.

Just a bit of caution, this formula can generate either a Primitive Pythagorean Triple or Imprimitive Pythagorean Triple. Remember that the former is a Pythagorean Triple where the Greatest Common Factor is equal to 1, while the latter has a GCF of greater than 1.

## The Pythagorean Triple Formula

**A few observations about the formula:**

- The variables m and n are any positive integers. However, positive integer m is greater than positive integer n, that is, m > n.

- We both use the values of m and n to calculate the lengths of the sides of right triangle ABC which are composed of the two legs a and b, and the longest side also known as the hypotenuse, represented by letter c.

- The value or length of side a is calculated by finding the difference of the squares of m and n. We can express it in the form of the equation as a = {m^2} - {n^2}.

- The length of side b is calculated by doubling the product of m and n. In the equation form, we have b = 2mn.

- Finally, the length of side c is computed by getting the sum of the squares of m and n. This can simply be written in the equation as c = {n^2} + {m^2}.

## Examples of Generating Pythagorean Triples

**Example 1:** Generate a Pythagorean Triple using the two integers 1 and 2.

First, observe that it is possible to generate a Pythagorean Triple with integers 1 and 2 because both are positive integers and one is larger than the other. We will let m be equal to 2 while n be equal to 1 because m should be greater than n based on the conditions above.

Since we now know what the values of **m** and **n**, it’s time to substitute those values into the formulas of a, b, and c to obtain the sides of the right triangle.

- Solving for a:

- Solving for b:

- Solving for c:

Let’s check if our values for a=3, b=4, and c=5 satisfy the Pythagorean Triple Equation which is {a^2} + {b^2} = {c^2}

Yes it does! Therefore, **(3,4,5)** is a Pythagorean Triple.

**Example 2:** Use the integers **3 **and **5** to generate a Pythagorean Triple. Is the generated triple a Primitive or Imprimitive Pythagorean Triple?

The problem above requires us to do two things. First, generate a Pythagorean Triple using the integers 3 and 5. Second, we need to figure out if the generated triple is Primitive or Imprimitive.

Check out my lesson on how to tell if a Pythagorean Triple is Primitive or Imprimitive.

◉ First part:

Since 5 is greater than 3, we have m=5 and n=3. Solving for a, b, and c, we have:

a = {m^2} - {n^2}

a = {(5)^2} - {(3)^2}

a = 25 - 9

a = 16

b = 2mn

b = 2\left( 5 \right)\left( 3 \right)

b = 30

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

c = {\left( 3 \right)^2} + {\left( 5 \right)^2}

c = 9 + 25

c = 34

Upon checking with the required formula, we verify that indeed **(16, 30, 34)** is a Pythagorean Triple.

◉ Second part:

Now, we want to know if our generated Pythagorean Triple is primitive or imprimitive.

Remember that a Pythagorean Triple is primitive if all three integers have a common factor of ONLY 1.

Otherwise, it is not primitive (imprimitive) since it has a common factor OTHER THAN 1.

By quick inspection, the Pythagorean Triple, (**16, 30, 34**), contains all even numbers. That means it has a factor other than 1 because all integers are divisible by 2. In other words, the triplet in question has more than 1 common factors. Therefore, (16, 30, 34) is an Imprimitive Pythagorean Triple.

**Example 3:** What is the Pythagorean Triple generated by the integers **3 **and **10**? Is the generated triplet primitive or non-primitive?

Since 10 > 3 then m = 10 and n = 3 because integer m must be larger than integer n .

At this point, you should already be familiar with the formula. Let’s solve for a, b, and c then.

- Computing for a :

- Solving for b :

- Calculating for c :

Verifying the values of a, b, and c with the Pythagorean Triple equation which is

\left( {a,b,c} \right) {a^2} + {b^2} = {c^2}we have,

\left( {91,60,109} \right) {91^2} + {60^2} = {109^2} {91^2} + {60^2} = {109^2} {8,281} + {3,600} = {11,881} {11,881} = {11,881}Yes! It checks. It means that \left( {91,60,109} \right) is indeed a Pythagorean Triple.

By quick inspection, we can tell that the integers **90**, **60**, and **109** have a Greatest Common Divisor of **1** which makes it a Primitive Pythagorean Triple.

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