Having trouble with these pairs of numbers (20,21,29) - (28,45,53) I am supposed to develop a Pythagorean triple. Here is that I am supposed to use which is a formula that can be used to generate infinite number of generating Pythagorean triples is:
a = (2mn); b= (m^2 - n^2); c= (m^2 + n^2)
where m and n are positive integers and m > n.
Here a example of how the book describe how it is suppose to look, but I still do not know how to figure out these two (20,21,29) - (28,45,53)Could someone help me?
Here is a example of how it should be done.
For m = 4, n = 3
a = 2mn = 2*4*3 = 24
b= m^2 - n^2 = 4^2 – 3^2 = 7
c= m^2 + n^2 = 4^2 + 3^2 = 25
Let us check (24, 7, 25) is a Pythagorean triple or not by using Pythagorean Theorem
a^2 + b^2 = 24^2 + 7^2 = 625
c^2 = 25^2 = 625
Since a^2 + b^2 = c^2 so (24, 7, 25) is a Pythagorean triple.
For m = 4, n = 1
a = 2mn = 2*4*1 = 8
b= m^2 - n^2 = 4^2 – 1^2 = 15
c= m^2 + n^2 = 4^2 + 1^2 = 17
20 = 2 m n
21 = m^2 - n^2
29 = m^2 + n^2
well, m n = 10
suspect m = 5 and n = 2 since those are factors of ten
m^2 = 25
n^2 = 4
sure enough 25 - 4 = 21
and
25+4 = 29
-------------------------
similarly
28,45,53
28 = 2 m n
so 14 = m n and m=7, n = 2 are factors
7^2= 49
2^2 = 4
49-4 = 45
49+4 = 53
so what is the answer?!
To generate a Pythagorean triple using the given formula, you need to choose values for m and n that satisfy the conditions m > n and both m and n are positive integers. Once you have chosen the values of m and n, you can substitute them into the formulas for a, b, and c.
Let's take the first pair of numbers (20, 21, 29) as an example. We need to find values for m and n that will satisfy the conditions.
Since a = 20, we can use the formula a = 2mn. Substituting the value for a, we have 20 = 2mn.
Now, let's try different values of m and n to see if we can find a solution. Starting with small values, we can choose m = 2 and n = 5. Substituting these values into the equation, we have 20 = 2 * 2 * 5, which is true.
Using the values of m = 2 and n = 5, we can find the values of b and c using the formulas b = m^2 - n^2 and c = m^2 + n^2. Substituting the values, we have b = 2^2 - 5^2 = 4 - 25 = -21 and c = 2^2 + 5^2 = 4 + 25 = 29.
The values (20, -21, 29) do not form a valid Pythagorean triple because b is negative.
Let's try a different set of values for m and n. We can choose m = 4 and n = 1. Substituting into the equation, we have 20 = 2 * 4 * 1, which is true.
Using m = 4 and n = 1, we can find b and c using the formulas b = m^2 - n^2 and c = m^2 + n^2. Substituting the values, we have b = 4^2 - 1^2 = 16 - 1 = 15, and c = 4^2 + 1^2 = 16 + 1 = 17.
The values (20, 15, 17) form a Pythagorean triple because a^2 + b^2 = 20^2 + 15^2 = 625 = 17^2.
Now, let's take the second pair of numbers (28, 45, 53) and follow the same steps.
Using m = 4 and n = 3, we can find a = 2 * 4 * 3 = 24, b = 4^2 - 3^2 = 16 - 9 = 7, and c = 4^2 + 3^2 = 16 + 9 = 25.
The values (24, 7, 25) form a Pythagorean triple because a^2 + b^2 = 24^2 + 7^2 = 625 = 25^2.
To generate more Pythagorean triples, you can continue to choose different values for m and n and apply the formulas.