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Suppose that $0\neq x\in\mathbb{R}$ and $x + \frac1x\in\mathbb{Z}$. Prove that, for all $n\ge1$, $x^n + \frac1{x^n}\in\mathbb{Z}$.

I can't figure out and understand the question. Can you give me some hints ?

comenman
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4 Answers4

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The base case of $n=1$ is true, and suppose it holds for all $k<n$ in order to do the induction step. Then $$(x^{n-1}+1/x^{n-1})(x+1/x)=x^n+1/x^{n-2}+x^{n-2}+1/x^n=(x^n+1/x^n)+(x^{n-2}+1/x^{n-2})$$ so $$x^n+1/x^n=(x^{n-1}+1/x^{n-1})(x+1/x)-(x^{n-2}+1/x^{n-2})$$ which is an integer, so the result follows by induction.

Matt Samuel
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  • @comenman You're welcome. – Matt Samuel Nov 29 '14 at 17:13
  • @MattSamuel Thanks for this answer. My question is would we not need to extend our base case to be allowed to use $x^{n-2}$ and $x^{n-1}$? I'm struggling a bit with strong induction but I thought that we needed to cover our minimal cases... that is if $n=1$ then $n - 2= k = -1$ which is outside of our claim... so would we not need to check up to $n=3$ in this case? – ClownInTheMoon Jan 02 '17 at 17:47
  • @Clown You're right, but $n=3$ isn't necessary. We can use $n=0$, in which case it's obvious. – Matt Samuel Jan 02 '17 at 18:35
  • @MattSamuel Yes I see that for $n=0$ it is obvious that the base case is true... but then how are we allowed to use $n-2$ for instance? Is this some application of vacuous truth? – ClownInTheMoon Jan 02 '17 at 19:01
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    @Clown For $n=2$ you use $n=1$ and $n=0$. For $n=3$, $n=2$ and $n=1$. You only need two initial terms for the recurrence. – Matt Samuel Jan 02 '17 at 19:06
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Hint: Expand $(x+\frac{1}{x})^n$ using the binomial theorem. Show that this is a linear combination of elements of the form $x^m+\frac{1}{x^m}$ for $m \leq n$ and of $1$. Then use induction.

Martin Brandenburg
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Hint let $$a_{n}=x^n+\dfrac{1}{x^n}$$ then we have $$a_{n+2}=(x+\dfrac{1}{x})a_{n+1}-a_{n}=ka_{n+1}-a_{n},x+\dfrac{1}{x}=k\in Z$$ since $a_{1}=k\in Z,a_{2}=k^2-2\in Z$ and it is by use Mathematical induction

math110
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The following is an extension to the hint provided by Martin Brandenburg: expanding $(1+1/x)^n$, which is an integer by hypothesis: $$nC0 * x^m * 1/x^0 + nC1 * x^(m-1) * 1/x + ... mC(m-1) * x * 1/x^(m-1) + mCm * x^0 * 1/x^m$$

Simplifying and rearranging, we get: $$x^m + x^(-m) + mx^(m-2) + mx^(2-m)... $$

Clearly, this is a linear combination of terms of the for $x^a + 1/x^a$ Now assume the original hypothesis is true for all k < n. Then the terms $$mx^(m-2) + mx^(2-m)...$$ are integers, leaving $$x^m + 1/x^m$$. However, since $(1+1/x)^n$ is an integer by hypothesis, so is $$x^m + 1/x^m$$.

Anonymous
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