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《数学通报》2010年11月号问题1882:在数列{an}中,a1=1,a2=4,当n≥3时,总有an=7an-1-an-2-2.求证:当n∈N*时,an均为平方数.
Problem 1882: In the series {an}, a1 = 1, a2 = 4, and when n≥3, there is always an = 7an-1-an-2-2. Proof: When n∈N *, an is the square number.