Induction: If $a+1/a$ is an integer, then so is $a^t+1/a^t$ for $t\in\mathbb N$

Question

$a$ is in $\mathbb{R}$ but not equal to $0$, and $a+\dfrac{1}{a}$ is integer, $a^t+\dfrac{1}{a^t}$ is also an integer for all $t\in\mathbb N$.

If $\displaystyle a+\frac1a$ is an integer then $\displaystyle \left(a+\frac1a\right)^2,\left(a+\frac1a\right)^3, \ldots $ are integers.

Maybe induction on $$a^{t+1} + \frac1{a^{t+1}} = \left(a^t+\frac1{a^t}\right)\left(a+\frac1a\right) - \left(a^{t-1}+\frac1{a^{t-1}}\right)$$

I'm having a problem doing the induction for this problem, any help will be appreciated.

You're basically done already. The right-hand side is a difference of two integers (by inductive hypothesis), so it's an integer. — Patrick Stevens, Mar 29 '17 at 21:57
@PatrickStevens Ah, but how do we know that $\left(a^t+\frac1{a^t}\right)$ is an integer? — , Mar 29 '17 at 22:02
HINT: If $a+1/a= b$ then $a^2+1/a^2=b^2-2$ and $a^3+1/a^3=b^3-3b.$ — DanielWainfleet, Mar 29 '17 at 22:17
Possible duplicate of Proving that $\frac{\phi^{400}+1}{\phi^{200}}$ is an integer. — martin.koeberl, Mar 29 '17 at 23:41
@GCab Set it equal to whatever integer you want and you get a quadratic equation in $k$ with real roots if $k\geq 2$. — martin.koeberl, Mar 29 '17 at 23:45
@GCab in the proof question, a is an element of Real numbers. — , Mar 29 '17 at 23:53

score 0 · Accepted Answer · answered Mar 29 '17 at 22:31

0

Your method is correct. The inductive hypothesis is "$(a^i+a^{-i})$ is an integer for all $i \leq t$". You will use this three times in each induction step: for $i=t$, for $i=1$, and for $i=t-1$.

answered Mar 29 '17 at 22:31

Patrick Stevens

36,135

Induction: If $a+1/a$ is an integer, then so is $a^t+1/a^t$ for $t\in\mathbb N$

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