Determine all real x in x^n+x^(-n)

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Discussion Overview

The discussion centers on determining all real values of \( x \) such that the expression \( x^n + x^{-n} \) is an integer for any integer \( n \). The scope includes mathematical reasoning and exploration of solutions derived from algebraic identities and properties of quadratic equations.

Discussion Character

  • Mathematical reasoning, Technical explanation

Main Points Raised

  • One participant suggests that if \( x + x^{-1} = k \) (where \( k \) is an integer), then \( x^2 + x^{-2} \) can be expressed as \( k^2 - 2 \), which is also an integer.
  • It is proposed that \( x^3 + x^{-3} \) can be derived from \( (x + x^{-1})^3 - 3(x + x^{-1}) \), leading to the conclusion that \( x^n + x^{-n} \) remains an integer for all integers \( n \).
  • The same participant concludes that it suffices to find all \( x \) such that \( x + x^{-1} = k \), which leads to a quadratic equation with solutions \( x = \frac{1}{2}(k \pm \sqrt{k^2 - 4}) \).
  • It is noted that these solutions are real for all integers \( k \) except for \( k = 0 \) or \( \pm 1 \).
  • The general solution is summarized as \( x = \frac{1}{2}(k \pm \sqrt{k^2 - 4}) \) for \( k \in \mathbb{Z} \) with \( |k| \geq 2 \).

Areas of Agreement / Disagreement

Participants appear to agree on the method of deriving the solutions and the conditions under which the solutions are valid, but there is no explicit consensus on the completeness or implications of the findings.

Contextual Notes

The discussion does not address potential limitations regarding the assumptions made about the values of \( k \) or the implications of the derived solutions for specific cases.

lfdahl
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Determine all real $x$ such that

$$x^n+x^{-n}$$

is an integer for any integer $n$

Source: Nordic Math. Contest
 
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–1, 0, and 1, because for any other whole numbers, $$x^{–n}$$ will become fraction while for fractions, $$x^n$$ will stay fraction.
 
Monoxdifly said:
–1, 0, and 1, because for any other whole numbers, $$x^{–n}$$ will become fraction while for fractions, $$x^n$$ will stay fraction.

zero is not valid. so x = -1 or 1
 
Monoxdifly said:
–1, 0, and 1, because for any other whole numbers, $$x^{–n}$$ will become fraction while for fractions, $$x^n$$ will stay fraction.

Hi, Monoxdifly

Your conclusion is wrong. Try e.g. the value: $x = 2 + \sqrt{3}$. (and $x=0$ is not valid as pointed out by kaliprasad)
 
lfdahl said:
Determine all real $x$ such that

$$x^n+x^{-n}$$

is an integer for any integer $n$

Source: Nordic Math. Contest
[sp]If $x+x^{-1} = k$ (where $k$ is an integer) then $x^2+x^{-2} = (x+x^{-1})^2 - 2 = k^2-2$, which is also an integer. Similarly, $x^3+x^{-3} = (x+x^{-1})^3 - 3(x+x^{-1})$ is also an integer, and, by an induction argument using the binomial theorem, $x^n+x^{-n}$ is an integer for all integers $n$.

So it is sufficient to find all $x$ such that $x+x^{-1} = k$. But that is a quadratic equation with solutions $x = \frac12\bigl(k \pm\sqrt{k^2-4}\bigr)$. Those solutions are real for all integers $k$ apart from $k = 0$ or $\pm1$.

Thus the general solution is $x = \frac12\bigl(k \pm\sqrt{k^2-4}\bigr)$ ($k\in\Bbb{Z},\ |k|\geqslant2$).[/sp]
 
Hi, Opalg!

Thankyou for your correct answer! I guess, I could save a lot of time writing down the suggested solution, whenever one of your exemplary solutions appears on the screen! :D
 

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