1

I have ben stuck on this for a while.

Let $F(N)$ be the Fibonacci numbers with $F(1)=F(2)=1$. Show that $4(-1)^n + 5(F(N))^2$ is a square for all integers $N$.

Casteels
  • 11,292
  • 4
  • 27
  • 38
MathFanatics
  • 335
  • 1
    Where are you stuck? – Masacroso Dec 30 '16 at 22:25
  • Presumably the exponent should be $N$, not $n$. They are not the same, though are often related. Have you done the base case? Then assume it is true up to $k$, plug in $k+1$ and play with what you get. Plug in the definition of $F(N+1)$. What happens? – Ross Millikan Dec 30 '16 at 22:25
  • I did check the base case byt I dont know how to prove it using My hypothesis – MathFanatics Dec 30 '16 at 22:30
  • Numerically, it looks like the square root of your expression satisfies the Fibonacci recursion, with $s_1=1,s_2=3$. I'd try to prove that inductively. – lulu Dec 30 '16 at 22:59
  • I'm sure this question has been asked and answered here before (but the closest I could find is http://math.stackexchange.com/questions/2032275/induction-proof-of-fibonacci-number which isn't quite the same thing). – Gerry Myerson Dec 31 '16 at 00:03
  • @lulu For reference, that sequence is known as the Lucas numbers $L_n$ A000204. So one evidently wants to show that $L_n =5F_n^2+4(-1)^n$. – Semiclassical Dec 31 '16 at 16:46
  • @Semiclassical Thanks. I've seen references to the Lucas numbers, but never looked them up. Apparently this connection to Fibonacci is well known, though it is new to me. – lulu Dec 31 '16 at 17:38

1 Answers1

1

This doesn't prove it inductively, so if you specifically need an inductive proof, this wouldn't work.

Instead, this uses the closed form for the Fibonacci sequence, which is that $F(N)=\dfrac{\alpha^N-\beta^N}{\sqrt{5}}$, where $\alpha=\frac{1+\sqrt{5}}{2}$ and $\beta=\frac{1-\sqrt{5}}{2}=\frac{-1}{\alpha}$.

The expression $4\cdot(-1)^N+5(F(N))^2$ becomes $$4\cdot(-1)^N+5\left(\dfrac{\alpha^N-\beta^N}{\sqrt{5}}\right)^2 =4\cdot(-1)^N+\alpha^{2N}-2\alpha^N\beta^{N}+\beta^{2N}.$$

Since $\beta=\frac{-1}{\alpha}$, $2\alpha^N\beta^N=2\cdot(-1)^N$ and so our expression becomes $$\begin{align} 4\cdot(-1)^N+\alpha^{2N}-2(-1)^{N}+\beta^{2N}= \\ \alpha^{2N}+2(-1)^N+\beta^{2N}= \\ \alpha^{2N}+2\alpha^N\beta^N+\beta^{2N}= \\ (\alpha^N+\beta^N)^2 \end{align}$$ which is a perfect square.

user402287
  • 204