MHB Challenge problem #2 Show that 5φ^2n+4(−1)^n is a perfect square

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Define a Fibonacci sequence by
$$\varphi_0=0,\,\varphi_1=1;\ \varphi_{n+2}=\varphi_{n+1}+\varphi_n\ \forall \,n\in\mathbb Z^+\cup\{0\}.$$
Show that
$$5\varphi_n^2+4(-1)^n$$
is a perfect square for all non-negative integers $n$.
 
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Olinguito said:
Define a Fibonacci sequence by
$$\varphi_0=0,\,\varphi_1=1;\ \varphi_{n+2}=\varphi_{n+1}+\varphi_n\ \forall \,n\in\mathbb Z^+\cup\{0\}.$$
Show that
$$5\varphi_n^2+4(-1)^n$$
is a perfect square for all non-negative integers $n$.
[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]
 
Opalg said:
[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]
Excellent solution! (Clapping) My own solution is much more mundane by comparison; I’ll wait and see if anyone else wants to try the problem before posting it.
 
Here’s my own solution to the challenge problem.
We shall prove by induction that the Fibonacci numbers satisfy the following equation:
$$\varphi_{n+1}^2-\varphi_n\varphi_{n+1}-\varphi_n^2-(-1)^n\ =\ 0.$$
It is easily checked that the equation is satisfied for $n=0$. Assume it is true for some integer $n\geqslant0$. Rewriting $\varphi_n=\varphi_{n+2}-\varphi_{n+1}$ gives
$$\varphi_{n+1}^2-(\varphi_{n+2}-\varphi_{n+1})\varphi_{n+1}-(\varphi_{n+2}-\varphi_{n+1})^2-(-1)^n\ =\ 0$$
which on simplifying becomes
$$-\varphi_{n+2}^2+\varphi_{n+1}\varphi_{n+2}+\varphi_{n+1}^2-(-1)^n\ =\ 0$$
– i.e.
$$\varphi_{n+2}^2-\varphi_{n+1}\varphi_{n+2}-\varphi_{n+1}^2-(-1)^{n+1}\ =\ 0.$$
QED. Hence: each Fibonacci number $\varphi_{n+1}$, an integer, is a root of the quadratic
$$x^2-\varphi_nx-\varphi_n^2-(-1)^n\ =\ 0$$
which has integer coefficients. Therefore its discriminant must be a perfect square. And the discriminant is
$$(-\varphi_n)^2-4\left[-\varphi_n^2-(-1)^n\right]$$
– that is to say:
$$5\varphi_n^2+4(-1)^n.$$
 
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