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Prove that if $x+\frac{1}{x}$ is an integer, then $x^n+\frac{1}{x^n}$ is also an integer.

I am aware this question has already been answered previously here and here, but I need help in proving it in an alternative way.

I am first trying to prove this for the case where $n$ is even.

Defining $$I_n:=x^n+\frac{1}{x^n},$$ I noticed that $$\\{\left(x+\frac{1}{x} \right)^n}=x^n+\frac{1}{x^n}+\sum_{k=1}^{n/2}{{n\choose k}{\left(x^{n-2k}+\frac{1}{x^{n-2k}}\right)}}$$ $$\Rightarrow I_n=(I_1)^n-\sum_{k=1}^{n/2}{{n\choose k}}\cdot I_{n-2k}$$

I initially planned on proving through induction, i.e., by proving that if $I_n$ is integer for some value of $n$, then $I_{n+2}$ must also be an integer, but for $I_{n+2}$ to be an integer, you also need $I_{n-2},I_{n-4}\ ,...I_{4},I_{2}$ to be integers. How do I mathematically prove that $I_{n}$ being an integer also implies all of these are also integers?

Moreover, would to be correct to instead assume that all $I_n, I_{n-2},I_{n-4}\ ,...I_{4},I_{2}$ are integers for some value of $n$ and then prove that $I_{n+2}$ is an integer? I ask this because I haven't seen induction proofs assuming the assumption to be true for more than 1 value of the variable.

William A. Noble
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Vulgar Mechanick
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3 Answers3

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For $n$ even, observe: $I_n = I_{n-2}\cdot I_2 - I_{n-4}$, [$1$]. Thus you can prove the claim using induction on $n$ being even. Starting with $n = 2$, $I_2 = \left(x+\frac{1}{x}\right)^2 - 2$ is an integer since $x+\dfrac{1}{x}$ is by assumption. Assume the claim is true for even $n$ up to $n-2$. This means $I_{n-2}, I_{n-4}$ are integers.Then [$1$] says that $I_n$ is also an integer. Thus by strong induction on even $n$, $I_n$ is an integer for all even $n$.

Wang YeFei
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  • This is the first time I'm reading about "strong" induction. Does that allow me to assume the identity to hold up to a given value of n instead of a single value? – Vulgar Mechanick Aug 08 '21 at 03:50
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    @OVERWOOTCH: It means the claim is true for all indices $k \le n - 2$, the value just before $n$ the one you are trying to prove. So it has to be true for a range of values prior to the one you need to verify to be true too. – Wang YeFei Aug 08 '21 at 03:53
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$$x^{n+1}+\frac1{x^{n+1}}=\left(x^n+\frac1{x^n}\right)\left(x+\frac1x\right)-\left(x^{n-1}+\frac1{x^{n-1}}\right).$$

Kenta S
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  • I have already seen this answer. I am trying to prove it through the binomial theorem instead as this fact doesn't come to me naturally at all. – Vulgar Mechanick Aug 08 '21 at 03:43
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    @WilliamA.Noble"I am aware this question has already been answered previously, but I need help in proving it in an alternative way." Did you even read the entire question? This is A proof by induction, but not the only one. – Vulgar Mechanick Aug 08 '21 at 03:45
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Generalizing, let $u_n =a^n+b^n $.

Then

$\begin{array}\\ u_1u_n &=(a+b)(a^n+b^n)\\ &=a^{n+1}+ab^n+ba^n+b^{n+1}\\ &=a^{n+1}+b^{n+1}+ab(b^{n-1}+a^{n-1})\\ &=u_{n+1}+abu_{n-1}\\ \text{so}\\ u_{n+1} &=u_1u_n-abu_{n-1}\\ \end{array} $

Therefore if $ab$, $u_1$ and $u_2$ are integers, then $u_n$ is an integer for $n \ge 1$.

Since $u_2 =a^2+b^2 =a^2+2ab+b^2-2ab =(a+b)^2-2ab =u_1^2-2ab $, if $a+b$ and $ab$ are integers, then $a^n+b^n$ is an integer for $n \ge 1$.

If $a+\frac1{a}$ is an integer, then $b = \frac1{a}$ so $ab=1$ is an integer, so $a^n+\frac1{a^n}$ is an integer for $n \ge 1$.

marty cohen
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