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

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The discussion focuses on proving that the expression 5φ²n + 4(-1)ⁿ is a perfect square for all non-negative integers n, using properties of the Fibonacci sequence defined by φ₀ = 0 and φ₁ = 1. The proof employs matrix representation and induction, showing that the determinant of the Fibonacci matrix leads to the relationship φₙ₋₁φₙ₊₁ - φₙ² = (-1)ⁿ. By manipulating the expressions, it is demonstrated that 5φ²n + 4(-1)ⁿ can be rewritten as a square of an integer, specifically (φₙ + 2φₙ₋₁)². The discussion concludes with an acknowledgment of the solution's elegance and invites further contributions.
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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 $n$.
 
Last edited:
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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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