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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