[sp]Let $A = \begin{bmatrix}0&1\\1&1\end{bmatrix} = \begin{bmatrix}\varphi_0&\varphi_1\\ \varphi_1&\varphi_2\end{bmatrix}$. By induction, $A^n = \begin{bmatrix}\varphi_{n-1}&\varphi_n\\ \varphi_n&\varphi_{n+1}\end{bmatrix}$. The inductive step is given by the calculation $$A^{n+1} = A^nA = \begin{bmatrix}\varphi_{n-1}&\varphi_n\\ \varphi_n&\varphi_{n+1}\end{bmatrix} \begin{bmatrix}0&1\\1&1\end{bmatrix} = \begin{bmatrix}\varphi_{n}&\varphi_{n-1} + \varphi_n\\ \varphi_{n+1}&\varphi_{n}+\varphi_{n+1}\end{bmatrix} = \begin{bmatrix}\varphi_{n}&\varphi_{n+1}\\ \varphi_{n+1}&\varphi_{n+2}\end{bmatrix}.$$ Since $\det A = -1$, it follows that $\det A^n = (-1)^n$. Therefore $\varphi_{n-1}\varphi_{n+1} - \varphi_{n}^2 = (-1)^n$. Then $$ \varphi_{n}^2 + (-1)^n = \varphi_{n-1}\varphi_{n+1} = \varphi_{n-1}(\varphi_{n} + \varphi_{n-1}) = \varphi_{n}\varphi_{n-1} + \varphi_{n-1}^2,$$ $$4\varphi_{n}^2 + 4(-1)^n = 4\varphi_{n}\varphi_{n-1} + 4\varphi_{n-1}^2, $$ $$5\varphi_{n}^2 + 4(-1)^n = \varphi_{n}^2 + 4\varphi_{n}\varphi_{n-1} + 4\varphi_{n-1}^2 = (\varphi_{n} + 2\varphi_{n-1})^2, $$ which is a perfect square since $\psi_{n} = \varphi_{n} + 2\varphi_{n-1}$ is an integer.
(The numbers $\psi_n$ are the
Lucas numbers.)[/sp]