Proving a properties of fibonacci numbers

[John112]

Let p[n] be the following property of Fibonacci numbers:
p[n]: f[itex]_{n+1}[/itex] * f[itex]_{n-1}[/itex] - (f[itex]_{n}[/itex])[itex]^{2}[/itex] = (-1)[itex]^{n}[/itex]. Prove p[n] by induction.

This is the proof I wrote. i used regular induction. Is weak induction sufficient to prove it or do I need to prove this by strong induction?

proof:

BASE STEP: p[1]: f[itex]_{2}[/itex] * f[itex]_{0}[/itex] - (f[itex]_{1}[/itex])[itex]^{2}[/itex] = (1 * 0) - 1 = -1

and (-1)[itex]^{1}[/itex] = -1

INDUCTIVE STEP:

assume P[k]: f[itex]_{k+1}[/itex] * f[itex]_{k-1}[/itex] - (f[itex]_{k}[/itex])[itex]^{2}[/itex] = (-1)[itex]^{k}[/itex]

show p[k+1]: f[itex]_{k+2}[/itex] * f[itex]_{k}[/itex] - (f[itex]_{k+1}[/itex])[itex]^{2}[/itex] = (-1)[itex]^{k+1}[/itex]
We know f[itex]_{k+2}[/itex] = f[itex]_{k+1}[/itex] + f[itex]_{k}[/itex]

p[K+1]: f[itex]_{k+2}[/itex] * f[itex]_{k}[/itex] - (f[itex]_{k+1}[/itex])[itex]^{2}[/itex] = (-1)[itex]^{k+1}[/itex]

substitute f[itex]_{k+2}[/itex] = f[itex]_{k+1}[/itex] + f[itex]_{k}[/itex] into p[k+1]

p[k+1]: (f[itex]_{k+1}[/itex] + f[itex]_{k}[/itex]) * f[itex]_{k}[/itex] - (f[itex]_{k+1}[/itex])[itex]^{2}[/itex] = (-1)[itex]^{k+1}[/itex]

= f[itex]_{k}[/itex] * f[itex]_{k+1}[/itex] + (f[itex]_{k}[/itex])[itex]^{2}[/itex] - (f[itex]_{k+1}[/itex])[itex]^{2}[/itex]

We know p[k]: f[itex]_{k+1}[/itex] * f[itex]_{k-1}[/itex] - (f[itex]_{k}[/itex])[itex]^{2}[/itex] = (-1)[itex]^{k}[/itex]

We can rewrite p[k] as: (f[itex]_{k}[/itex])[itex]^{2}[/itex] = f[itex]_{k+1}[/itex] * f[itex]_{k-1}[/itex] - (-1)[itex]^{k}[/itex]

We substitue (f[itex]_{k}[/itex])[itex]^{2}[/itex] from p[k] into the (f[itex]_{k}[/itex])[itex]^{2}[/itex] in p[k+1] = f[itex]_{k+1}[/itex] + f[itex]_{k}[/itex] + f[itex]_{k+1}[/itex] * f[itex]_{k-1}[/itex] - (-1)[itex]^{k}[/itex] - f[itex]_{k+1}[/itex][itex]^{2}[/itex])

= f[itex]_{k+1}[/itex] (f[itex]_{k+1}[/itex] + f[itex]_{k}[/itex]) (-1)[itex]^{k}[/itex] - f[itex]_{k+1}[/itex][itex]^{2}[/itex])

= (f[itex]_{k+1}[/itex])[itex]^{2}[/itex]) - (-1)[itex]^{k}[/itex] - (f[itex]_{k+1}[/itex])[itex]^{2}[/itex])

= - ((-1)[itex]^{k}[/itex])

= (-1)[itex]^{k+1}[/itex]Does this proof strong enough or would I have to do this by strong induction? If by strong induction, how do I got about proving it?

 

[mfb]

Every valid proof with weak induction is a valid proof with strong induction, too (and this is the first time I saw specific words for them). Your proof uses valid steps only, therefore it is valid. There is no need to do more.