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Determine the remainder of the Pell number $$P_{gcd(\phi{6640} , \phi{4648})} $$ divided by 4

$$ gcd(\phi_m ,\phi_n)=\phi_{gcd(m,n)} \Rightarrow{} gcd(\phi{6640} , \phi{4648})=\phi_{664} $$

Because F(664) is of the form F(4n) then is divisible by 3

I don´t know how to continue or if it´s correct my approach

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    If $3 | F(4n)$, then $F(4n) mod 3$ is congruent to $0$, hence has $0$ remainder. However how do you know that $3 | F(4n)$? – 2017-01-19

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