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Giả sử: gcd(kFn+2+Fn;kFn+3+Fn+1)=d
⇒{d|kFn+3+Fn+1−kFn+2−Fn=kFn+1+Fn−1d|kFn+3+Fn+1+kFn+2+Fn=kFn+4+Fn+2
Bằng quy nạp suy ra: d|kFn+2+Fn,∀n∈N
Suy ra:
{d|kF2+F0=2k+1d|kF3+F1=3k+1⇒{d|6k+3d|6k+2
⇒d|(6k+3)−(6k+2)=1⇒d=1.
Suy ra: kFn+2+FnkFn+3+Fn+1 tối giản.
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