Let $a$ and $b$ is roots of the $x^2-6x+1$ equation. a) Show that, for all $n \in{\mathbb Z^+}$ , $a^n+b^n$ is a integer. b) Show that, for all $n \in{\mathbb Z^+}$ , $5$ isn't divide $a^n+b^n$
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Tags: number theory, number theory unsolved
joa51736
31.01.2016 14:32
a) The roots of the polynomial $x^2-6x+1$ are $3-2\sqrt{2}$ and $\frac{1}{3-2\sqrt{2}}$. Then it's trivial by the following lemma : Let $x$ be a positive real number such that $x+\frac{1}{x}$ is an integer. Then for all positive integers $n$, $x^n+\frac{1}{x^n}$ is an integer.
IsoLyS
31.01.2016 14:33
In fact $x_{n+1}=6x_{n}-x_{n-1}$ where $x_{n}=a^n+b^n$.
joa51736
31.01.2016 14:41
b) Use induction. Hint : $(x+\frac{1}{x})(x^n+\frac{1}{x^n})=x^{n+1} + \frac{1}{x^{n+1}} + x^{n-1} + \frac{1}{x^{n-1}}$.
GuvercinciHoca
06.05.2022 14:37
Define the sequence $\big( a_n\big )$ with $a_{n+2}=6a_{n+1} - a_n$ and $a_0=2, a_1=6.$ Now we have $a_n= a^n+b^n$ which obviously is an integer. Also we can manually check that $5$ never divides terms of this sequence.