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Is the following Proof Correct? In particular please comment on the correctness of the given formulas.

Theorem. Given that $x$ is a real number, $x\neq 0$, and $x + \frac{1}{x}$ is an integer. For all $n\ge 1$, $x^n+\frac{1}{x^n}$ is an integer.

Proof. We construct the proof by recourse to Strong-Induction. Assume for an arbitrary $n\in\mathbf{Z^+}$ that $x^k+\frac{1}{x^k}$ is an integer for any positive integer $k$ strictly less than $n$. Now Consider the following cases.

Case-1: If $n$ is even then for some $l\in\mathbf{Z^+}$, $n = 2l$ thus the Binomial-Theorem implies that $$(x+\frac{1}{x})^n - \sum_{j=1}^{l-1}\binom{n}{2j}\left(x^{2j}+\frac{1}{x^{2j}}\right)-\binom{n}{l} = x^n+\frac{1}{x^n}$$ from the inductive hypothesis we know that $x+\frac{1}{x}\in\mathbf{Z}$ which implies that $(x+\frac{1}{x})^n \in\mathbf{Z}$ in addition it also follows from the inductive hypothesis that $x^{2j}+\frac{1}{x^{2j}}\in\mathbf{Z}$ for $j\in\{1,2,3,...,l-1\}$ moreover we also that $\binom{n}{r}\in\mathbf{Z^+}$ is always a positive integer implying that $x^n+\frac{1}{x^n}$ is an integer.

Case-2: If $n$ is odd then for some $l\in\mathbf{Z^+}$, $n = 2l+1$ thus the Binomial-Theorem implies that $$(x+\frac{1}{x})^n - \sum_{j=1}^{l}\binom{n}{2j-1}\left(x^{2j-1}+\frac{1}{x^{2j-1}}\right) = x^n+\frac{1}{x^n}$$ and by using the same reasoning as in the previous case we can deduce that $x^n+\frac{1}{x^n}$ is an integer.

atifcppprogrammer
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  • @DietrichBurde why are you ignoring the proof verification tag? The OP is not asking for a proof but for a review of his efforts. – Thomas Jan 17 '18 at 19:16
  • It seems good to me. But a bit to complicated. Let $a_i = x^i + \frac 1{x^i}$ be an integer for $1 \le i \le n$. Then $a_1a_n = (x^n + \frac 1{x^n})(x + \frac 1x) = $$x^{n+1} + x^{n-1} + \frac 1{x^{n-1}} + \frac 1{x^{n+1}} = $$x^{n+1} + \frac 1{x^{n+1}} + a_{n-1}$. So $x^{n+1} + \frac 1{x^{n-1}} = a_1a_n - a_{n-1}$ is an integer. But to the degree of which I read yours, it looks good (in fact it's basically the same thing I did). – fleablood Jan 17 '18 at 19:16
  • @Thomas Because the OP should also compare the proofs for duplicate questions, and then can specifically ask about it. – Dietrich Burde Jan 17 '18 at 19:18
  • @DietrichBurde But then your link does not apply – Thomas Jan 17 '18 at 19:19
  • I'll admit, I didn't read the OP's proof in great detail, but I can assure the OP that if s/he has done it correctly, that method will work. (I can also suggest it's a bit too complicated) – fleablood Jan 17 '18 at 19:21

2 Answers2

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I get the gist of the argument and it looks correct to me. To verify it, though, I had to stare at the summations in the middle of the two displayed equations for a while.

This would be annoying to a grader (graders are lazy, stupid, and mean). So I suggest inserting your steps before those equations:

  1. Expand $\left(x + \frac{1}{x}\right)^{2l}$ with the Binomial Theorem
  2. Visually collect the terms into multiples of $\left(x^{k} + \frac{1}{x^{k}}\right)$
  3. Apply the inductive hypothesis.

Basically I am asking for you to slow down at the point "The Binomial Theorem implies that" and be more explicit.

Matthew Leingang
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2

I think it is easier to see that

$$ x^{n+2}+\frac{1}{x^{n+2}}=\left(x+\frac{1}{x}\right)\left(x^{n+1}+\frac{1}{x^{n+1}}\right)-\left(x^n+\frac{1}{x^n}\right) $$

Atmos
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