Let S₁=S₂=1, S₃=4, and S_n+3= 2S_n+2+2S_n+1-S_n for n≥1.

Is S_p always a perfect square?

S(n)=F(n)^2, the n-th Fibonacci number squared. To see why, we can substitute in the equation for S(n+3) getting:

F(n+3)²=2F(n+2)²+2F(n+1)²-F(n)²

...=F(n+2)²+2F(n+2)F(n+1)+F(n+1)²
     +F(n+2)²-2F(n+2)F(n+1)+F(n+1)²-F(n)²

...=(F(n+2)+F(n+1))²+(F(n+2)-F(n+1))²-F(n)²

...=F(n+3)²+F(n)²-F(n)²

...=F(n+3)²

QED

Posted by Old Original Oskar! on 2007-03-14 14:49:54