question may arise sama ada sesuatu integer positif z {\displaystyle z} is a nombor Fibonacci. Since F ( n ) {\displaystyle F(n)} is closest integer to φ n / 5 {\displaystyle \varphi ^{n}/{\sqrt {5}}} , most straightforward, brute-force test is identity

F ( ⌊ log φ ⁡ ( 5 z ) + 1 2 ⌋ ) = z , {\displaystyle F{\bigg (}{\bigg \lfloor }\log _{\varphi }({\sqrt {5}}z)+{\frac {1}{2}}{\bigg \rfloor }{\bigg )}=z,}

which is benar jika dan hanya sekiranya z {\displaystyle z} merupakan nombor Fibonacci.

Alternatively, a integer positif z {\displaystyle z} ialah nombor Fibonacci sekiranya dan hanya sekiranya salah satu 5 z 2 + 4 {\displaystyle 5z^{2}+4} or 5 z 2 − 4 {\displaystyle 5z^{2}-4} merupakan segi empat yang sempurna.[10]

A slightly more sophisticated test uses fact that convergents perwakilan pecahan berterusan φ {\displaystyle \varphi } ialah nisbah-nisbah nombor Fibonacci yang berturutan, that is inequality

| φ − p q | < 1 q 2 {\displaystyle {\bigg |}\varphi -{\frac {p}{q}}{\bigg |}<{\frac {1}{q^{2}}}}

(with coprime integer positif p {\displaystyle p} , q {\displaystyle q} ) is benar if and only if p {\displaystyle p} and q {\displaystyle q} are successive nombor Fibonacci. From this one derives criterion that z {\displaystyle z} is a nombor Fibonacci if and only if closed interval

[ φ z − 1 z , φ z + 1 z ] {\displaystyle {\bigg [}\varphi z-{\frac {1}{z}},\varphi z+{\frac {1}{z}}{\bigg ]}}

contains a integer positif.[11]