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Algebra Difficulty 4.8 AIME Prove it Brazil

The sequence a1,a2,a3,a_1, a_2, a_3, \ldots is defined by a1=8a_1 = 8, a2=18a_2 = 18, an+2=an+1ana_{n+2} = a_{n+1} \cdot a_n.
Find all terms which are perfect squares.

Solution

We have an=2bn3cna_n = 2^{b_n} 3^{c_n}, where bn+2=bn+bn+1b_{n+2} = b_n + b_{n+1}, and cn+2=cn+cn+1c_{n+2} = c_n + c_{n+1}.

c1c_1 and c2c_2 are even, so all cnc_n are even.

b1b_1 and b2b_2 are odd, so bnb_n is even for nn multiple of 33.

We need both bnb_n and cnc_n even for ana_n to be a perfect square.

So the answer is all nn multiple of 33.

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