This article deals with Pythagorean Triples and integer solutions of it. I want to find new things in mathematics. I find something then I search for it on internet whether it is already found or not. Few months back I wanted to find integer solutions of Pythagorean triples. I found a method which I will give in this article. This method uses algebra and ratio and proportion to solve the problem. The method derives two linear equations with the help of the equation x
^{2} + y^{2} = z^{2}. At the end we find an identity. The identity gives us three numbers which can be used to produce Pythagorean triples. The first three numbers which are Pythagorean triples are 3,4 and 5.
Consider the equation:
We can find odd Pythagorean triples by taking k = 2n where n is the number. Then dividing all the Pythagorean triple numbers by 2. We can find many properties with the help of this relation. One I have already used. When all the Pythagorean triples are multiplied by two then the numbers formed are also Pythagorean triples.
x^{2} + y^{2} = z^{2}⇒ x^{2} = z^{2} − y^{2}⇒ x^{2} = (z + y)(z − y)⇒ x/(z + y) = (z − y)/xLet it be equal to k. Then ⇒ x/(z + y) = (z − y)/x = kThe two simultaneous equations obtained from the above equality are x/(z + y) = k gives x = zk + ykand (z − y)/x = k gives kx = z − ySolving the above two linear equations, we get x/y = 2k/(1 − k^{2}) andx/z = 2k / (1 + k^{2})x: y : z = 2k : (1 − k^{2}):(k^{2} + 1)Substituting in x we get^{2} + y^{2} = z^{2}(2k)^{2} + (1 − k^{2})^{2} =(k^{2} + 1)^{2}(2k)
^{2} + (−(k^{2} − 1))^{2} =(k^{2} + 1)^{2} (2k)
^{2} + (k^{2} − 1)^{2} =(k^{2} + 1)^{2}the above equation is an identity, when we expand it we obtain it. When we take k as any numberthen the three integer Pythagorean triples are 2k, (k^{2} − 1) and (k^{2} + 1) |

My discoveries > Diophantine Equations >