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In order to find a shorter proof for this thread :

Find all functions $f(m)[(f(n))^2-1]=f(n)[f(m+n)-f(m-n)]$

Rem: just to be clear, $\mathbb N^*$ stands for $\mathbb N\setminus\{0\}$


I'm interested in whether it is possible to establish directly that

If $\ \forall n>0$ then $F_n=\left(\alpha\,a^n + \beta\, \dfrac{(-1)^n}{a^n}\right)\in\mathbb N^*\quad$ with $(\alpha,\beta)\in\mathbb R^2$ and $a\in\mathbb N^*$

$\implies (a=1) \text{ or }(\beta=0)$.


A possible additional condition could be to set $F_1=a$ also.

This seems expected because intuitively $\beta$ has to be divisible by any $a^n$, but when trying to prove it, I only manage to show that $\alpha,\beta$ are rational, but cannot find the decisive blow.

I'm sure I'm missing something obvious, but I can't see it...

zwim
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2 Answers2

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Note that $a \alpha = F_1 + \frac{\beta}{a},$ which implies that $$F_n = F_1 a^{n-1} + \beta \left( \frac{a^{2n-2} +(-1)^n}{a^n} \right).$$ Since both $F_n$ and $F_1 a^{n-1}$ are naturals for every $n$, we must have that for every $n$, $$\beta \left( \frac{a^{2n-2} + (-1)^n}{a^n} \right) \in \mathbb{Z}.$$ In particular, this would hold for all odd $n$. Now note that for every $a, n >1 \in \mathbb{N},$ $a^n$ is relatively prime to $a^{2n-2} - 1.$ Indeed, if $d$ divides both $a^n$ and $a^{2n-2} - 1,$ then it must also divide $a^{n-2} a^n - (a^{2n-2} - 1) = 1,$ forcing $d = 1$.

Thus, there are infinitely many $n$ such that $\beta /a^n$ is an integer. But, since $a>1,$ for large enough $n$ we have $|\beta/ a^{n}| <1, $ forcing $\beta = 0$.

stochasticboy321
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  • Very good answer, and explanation! +1 – Fimpellizzeri Oct 17 '17 at 02:26
  • The expression of $F_n$ in function of $F_1$ is a nice idea to make $\alpha$ out of the equation, this is indeed what I missed, could not figure out how to dismiss $\alpha a^n$ and $\beta a^{-n}$ at the same time. – zwim Oct 17 '17 at 15:37
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No. It is fairly common.

An example is the Fibonacci numbers.

(see https://en.wikipedia.org/wiki/Fibonacci_number#Relation_to_the_golden_ratio)

They satisfy

$F_n =\dfrac1{\sqrt{5}}(\phi^n-(-\phi)^{-n}) $ where $\phi = \dfrac{1+\sqrt{5}}{2} $.

This is because the $\phi$ satisfies $x^2-x-1 = 0$.

This will happen for recurrences of the form $a_{n+2} = ka_{n+1}+a_n$.

marty cohen
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