1F (Numbers and Sets)

The two conjugate roots satisfy

r^{2} = 6 r - 4

, so

u_{n} = 6 u_{n - 1} - 4 u_{n - 2}

with

u_{0} = 2, u_{1} = 6

. Induction gives

u_{n} \in Z

ii

The same recurrence and

2^{0} ∣ u_{0}

2^{1} ∣ u_{1}

prove the claim: if

2^{n - 1} ∣ u_{n - 1}

and

2^{n - 2} ∣ u_{n - 2}

, both terms on the right are divisible by

2^{n}

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