Putting everything in terms of just two of the numbers and expanding both sides (of works.
and
Substitute and expand both sides, they turn out to be identical.
The problem is as follows:
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We all know that this is the Fibonacci recusion.
F_{n }= F_{n-1 }+ F_{n-2} , F_{0 }= 1 , F_{1} = 1
How to prove that the following triple (x,y,z) follow Pythagorean Theorem ?
x = (F_{i}) x (F_{i+3})
y = 2 x (F_{i+1}) x (F_{i+2})
z = (F_{i+1})^{2} + (F_{i+2})^{2}
Thanks but I've found another simple proof for this...
F_{i+3} = F_{i+2} + F_{i+1}
and
x = F_{i} x F_{i+3} = (F_{i+2} - F_{i+1}) x (F_{i+2} + F_{i+1}) = b^{2} - a^{2} where
b = F_{i+2} and a = F_{i+1}.
Now (b^{2} - a^{2})^{2} + (2ab)^{2} = (b^{2} + a^{2})^{2}
Hence proved...
Anyways thanks for your proof too...