Friday, October 5, 2012

How do you calculate Pythagorean triples with 2 consecutive integers?Also, how do you find a pattern to generate infinitely many Pythagorean...

To generate the pythagorean triplets with consecutive
integers.


We know any three positive numbers of the type a,
b and c  are a Pythagorean if a^2+b^2 = c^2.


but our
interest is to generate Pythagorean triplets.


We know  (a^2
+x^2 ) ^2 - (a^2-x^2)^2 = 4a^2x^2 is an identity valid for all a and
x.


Or (a^2+x^2)^2 - (a^2-x^2) = (2ax)^2.
Or


(a^2-x^2)^2 + (2ax)^2 = (a^2+x^2)^2  is an identity so
it is true for all x and a.


Therefore (a^2-x^2) , 2ax and
(a^2+x^2) is a Pythagorean triplet for all a and
x.


Example: a= 13, x = 4. Then a^2-x^2 = 13^2 -4^2 =
153


2ax = 104 and a^2+x^2 = 
185.


Therefore 153^2+104^2 = 34225.  but 185^2 =
34225.


Now to get successive integers as
triplets:


We know (a^2+x^2)-(a^2-x^2) =
4a^2x^2....(1)


We put x^2 = 1^2 in
(1):


(a^2+1)^2-(a^2-1)^2 =
4a^2.


Divide by
4:


{(a^2+1)/}^2 - {a^2-1)/2}^2 =
a^2.


Therefore  a^2 + {a^2-1)/2}^2 = {a^2
+1)/2}^2.


Now we can generate two  consecutive integers and
another integer as a Pythagorean triplet from the
triplet:


a , (a^2-1)/2 and (a^2+1)/2. But  we have to take
a as an odd integer.


Example; a = 25. Then (a^2-1)/2 =
(25^2-1)/2 = 624/2 = 312.  (a^2+1)/2 = (626/2 = 323.


25^2
+312^2 = 97969


313^2 =
97969.


Therefore 25, 312 and 313 are the Pythagorean
triplet with 312 and 313 as consecutive integers.


Example
2: a = 7. (a^2-1)/2 = (7^2-1)/2 = 24 , (a^2+1)/2- (7^2+1)/2 =
25.


7^2+24^2 = 625 which is
25^2.


Therefore 7,24 and 25 are Pythagorean triplet with 24
and 25 as cosecutive
integer.



T

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