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

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In summary, the problem is to determine all real $x$ such that $x^n+x^{-n}$ is an integer for any integer $n$. The solution involves finding the general solution for the quadratic equation $x+x^{-1}=k$, which is $x = \frac12\bigl(k \pm\sqrt{k^2-4}\bigr)$ ($k\in\Bbb{Z},\ |k|\geqslant2$). This solution applies for all integers $n$ apart from $n=0$ or $\pm1$.
  • #1
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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  • #2
–1, 0, and 1, because for any other whole numbers, \(\displaystyle x^{–n}\) will become fraction while for fractions, \(\displaystyle x^n\) will stay fraction.
 
  • #3
Monoxdifly said:
–1, 0, and 1, because for any other whole numbers, \(\displaystyle x^{–n}\) will become fraction while for fractions, \(\displaystyle x^n\) will stay fraction.

zero is not valid. so x = -1 or 1
 
  • #4
Monoxdifly said:
–1, 0, and 1, because for any other whole numbers, \(\displaystyle x^{–n}\) will become fraction while for fractions, \(\displaystyle x^n\) 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)
 
  • #5
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]
 
  • #6
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
 

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

1. What is the equation x^n+x^(-n)?

The equation x^n+x^(-n) is a polynomial expression with two terms, where n is a positive integer. The first term, x^n, is the base raised to the power of n, while the second term, x^(-n), is the reciprocal of the base raised to the power of n.

2. How do you determine the real solutions for x in x^n+x^(-n)?

To determine the real solutions for x in x^n+x^(-n), we can use the laws of exponents and algebraic manipulation to rewrite the equation as (x^n)^2 + 1 = 0. Then, we can use the quadratic formula to solve for x, which will give us two real solutions.

3. Can x be any real number in x^n+x^(-n)?

No, x cannot be any real number in x^n+x^(-n). The equation is only defined for positive integers n. Additionally, if n is an even number, x must be a positive real number to avoid taking the square root of a negative number.

4. Are there any restrictions on the values of n in x^n+x^(-n)?

Yes, there are restrictions on the values of n in x^n+x^(-n). As mentioned before, n must be a positive integer. Additionally, if n is an even number, the equation will only have real solutions if x is a positive real number.

5. Can x^n+x^(-n) have complex solutions?

Yes, x^n+x^(-n) can have complex solutions. If n is an odd number, the equation will have two complex solutions. If n is an even number and x is a negative real number, the equation will have two complex solutions as well.

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