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The recurrence relation seems pretty messy, so I will try to solve it inductively, T_T

For n=0.

I would suggest induction as bmp is attempting.

@bmp, check your value for \(a_7\)

Ah, got it, I think.

Thanks @KingGeorge :-)

Fibonacci it is, then :-)

Fibonacci How?

I think every
\[a_{n} = Fib(n)*Fib(n)\]that is a perfect square for every n in N.

But how are you going to prove it?

to prove*

Another hint that may be helpful:
Use strong induction.

Yeah, I just now got to it. I think that would be the way, if you can prove for a(1)...a(k).

@m_charron2 Kudos for beating me on the proof, mate :-). Well done.

For those interested, here's the solution I got when I was challenged with this problem.

on this note, off to sleep for me, that drained me lol. Excellent problem KingGeorge!

Good night.