# How to find pythagoras triplet using the fibonacci sequence?

I'm using the Fibonacci sequence to generate some Pythagorean triples $(3, 4, 5,$ etc$)$ based off this page:Formulas for generating Pythagorean triples starting at "Generalized Fibonacci Sequence".

For Fibonacci numbers starting with $F_1=0$ and $F_2=1$ and with each succeeding Fibonacci number being the sum of the preceding two, one can generate a sequence of Pythagorean triples starting from $(a_3, b_3, c_3) = (4, 3, 5)$ via $$(a_n, b_n, c_n) = (a_{n-1}+b_{n-1}+c_{n-1}, F_{2n-1}-b_{n-1}, F_{2n})$$

for $n \ge 4$.

I am unable to generate Pythagorean triplet sequence using Fibonacci series.

Kindly Help!!!!!!!!!

You should get something like

n   Fib_{2n-1}  Fib_{2n}    a_n b_n c_n
3       3           5       4   3   5
4       8           13      12  5   13
5       21          34      30  16  34
6       55          89      80  39  89
7       144         233     208 105 233
8       377         610     546 272 610


etc.

• But i have to generate pythagoraes triplet ,, where is the case {6,8,10} etc. @Henry – Aditya Sharma Oct 3 '15 at 18:06
• @AdityaSharma: Try to calculate $a_n^2+b_n^2-c_n^2$ from the values in my table. It generates some Pythagorean triplets, not all of them. – Henry Oct 3 '15 at 18:07
• where is the case {6,8,10},etc.... in your case a_n^2+b_n^2-c_n^2=0 – Aditya Sharma Oct 3 '15 at 18:10
• some not all – Henry Oct 3 '15 at 18:11
• but i want to generate all triplets , how to generate them? – Aditya Sharma Oct 3 '15 at 18:12