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Let it be $a \neq 0,a \in \mathbb{R} $ such that $a+(1/a) \in \mathbb{Z} $. prove by induction that for all $n \in \mathbb{N}: a^n +(1/a^n)\in \mathbb{Z} $

i tried simplifying the phrase but couldn't use the given claim to prove something

ga as
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    Do you mean "such that $a+(\frac{1}{a}) \in \Bbb Z$"? If not, the claim (with $\Bbb R$ instead of $\Bbb Z$) is false. Take $n=1$ and $a=\frac{1}{2}$, for example. – xyz Nov 19 '22 at 14:47
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    Hint: $$a^{n+1}+\frac{1}{a^{n+1}} = \left(a^n+\frac{1}{a^n}\right)\left(a+\frac{1}{a}\right) - \left(a^{n-1}+\frac{1}{a^{n-1}}\right).$$ – Riemann Nov 19 '22 at 14:49
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    yes i ment Z, thanks so much! since all of the three phrases in Z that the result is also in Z – ga as Nov 19 '22 at 14:54
  • While not a proof by induction, another way to prove this is via the fundamental theorem of symmetric polynomials. Since $a + \frac{1}{a}$ is an integer and $a \times \frac{1}{a}=1$ is an integer, any symmetric polynomial in terms of $a$ and $\frac{1}{a}$ must be an integer – Aditya Gupta Nov 19 '22 at 15:01
  • @roro could you please elaborate how did you come up with simplification? I want to be able to do this alone next time i see this kind of questions – ga as Nov 19 '22 at 15:15
  • @gaas I wasn't the one to come up with the simplification. It is explained in Feripinho answer thought. – xyz Nov 19 '22 at 15:17

1 Answers1

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Let $a\in\mathbb{R}^*$, such that $a + \dfrac{1}{a}\in\mathbb{Z}$. If $n\in\mathbb{N}$, so $a^n + \dfrac{1}{a^n}\in\mathbb{Z}$.

  1. If $n = 1$, then $a^1 + \dfrac{1}{a^1} = a + \dfrac{1}{a} \in\mathbb{Z}$;
  2. Suppose it also works until $n = p$: $a^p + \dfrac{1}{a^p}\in\mathbb{Z}$, that is; then
  3. We show it also works for $n = p + 1$. Indeed, we multiply the expressions for $n = 1$ and for $n = p$. Since they're both integers, their product is also an integer: $\left(a + \dfrac{1}{a}\right)\left(a^p + \dfrac{1}{a^p}\right) = a^{p+1} + \dfrac{1}{a^{p+1}} + a^{p-1} + \dfrac{1}{a^{p-1}}\in\mathbb{Z}$. By step 2, $a^{p-1} + \dfrac{1}{a^{p-1}}\in\mathbb{Z}$ (since $p - 1 < p$ and it works until $p$). Therefore, we subtract from step 3's expression this last one, getting also an integer, since the subtraction of two integers is also an integer. Hence, $a^{p+1} + \dfrac{1}{a^{p+1}}\in\mathbb{Z}$.

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\square$

Feripinho
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  • could you please elaborate how did you come up with simplification? I want to be able to do this alone next time i see this kind of questions? how did you know that in order to solve it we need to multiply the expression of n=1 and the expression n=p?? – ga as Nov 19 '22 at 15:18
  • That's actually a great question. There's no reason why. I knew I had to reach $a^{p+1} + \dfrac{1}{a^{p+1}}$ somehow, so it sounded reasonable to multiply that expressions. It was basically faith. – Feripinho Nov 19 '22 at 15:28