Mathematical Induction on Fibonacci numbers 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 $N$.

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.
