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

• MHB
• Olinguito
In summary: Let $A = \begin{bmatrix}0&1\\1&1\end{bmatrix} = \begin{bmatrix}0&0\\1&1\end{bmatrix}$. Then 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 $$A^{n+1} = A^nA = \begin{bmatrix}\varphi_{n-1}&\varphi_n\\ \varphi_n&\var Olinguito Define a Fibonacci sequence by$$\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$$is a perfect square for all non-negative integers n. Last edited: Olinguito said: Define a Fibonacci sequence by$$\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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## 1. How do you approach proving that 5φ^2n+4(−1)^n is a perfect square?

To prove that an expression is a perfect square, we need to show that it can be written as the square of an integer. This can be done through various approaches such as algebraic manipulation, using number theory, or proving by induction.

## 2. What is the significance of the values 5 and 4 in the expression 5φ^2n+4(−1)^n?

The values 5 and 4 are significant because they are the coefficients of the terms φ^2n and (−1)^n respectively. These coefficients play a crucial role in determining the properties of the expression and ultimately proving that it is a perfect square.

## 3. Can you explain the role of φ (phi) in this challenge problem?

φ (phi), also known as the golden ratio, is a mathematical constant with various interesting properties. In this particular challenge problem, φ^2n represents a perfect square, making it an essential component in proving that the entire expression is also a perfect square.

## 4. Is there a specific formula or method to prove that an expression is a perfect square?

No, there is no specific formula or method to prove that an expression is a perfect square. The approach may vary depending on the given expression and may require some creativity and critical thinking. However, some common techniques used in such proofs include completing the square, factoring, and using the properties of perfect squares.

## 5. Can you provide an example of a similar challenge problem that proves an expression to be a perfect square?

One possible example is proving that 5n^2 + 10n + 6 is a perfect square for all positive integers n. This can be done by showing that it can be written as (n+1)^2, making it a perfect square. The proof involves using the properties of perfect squares and algebraic manipulation.

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