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
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
2 answers
so what is the answer?!