# 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.