# Primes...

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• Aug 6th 2010, 01:25 PM
Also sprach Zarathustra
Primes...
Given that $\displaystyle x+\frac{1}{x}$ is a prime number.

Can $\displaystyle x^3+\frac{1}{x^3}$ also be a prime number?
• Aug 6th 2010, 02:57 PM
Bruno J.
Quote:

Originally Posted by Also sprach Zarathustra
Given that $\displaystyle x+\frac{1}{x}$ is a prime number.

Can $\displaystyle x^3+\frac{1}{x^3}$ also be a prime number?

Let $\displaystyle a=x^3+\frac{1}{x^3}$ and $\displaystyle p=x+\frac{1}{x}$.

We have $\displaystyle p^3 = a+3p$, hence $\displaystyle p | a$, hence no! (Wink)
• Aug 6th 2010, 03:19 PM
Also sprach Zarathustra
One point to Bruno J.
• Aug 6th 2010, 05:58 PM
simplependulum
except the case $\displaystyle x = 1$ (Happy)

$\displaystyle p^2 - 3$ may equal $\displaystyle 1$
• Aug 6th 2010, 07:34 PM
Soroban
Hello, Also sprach Zarathustra!

Quote:

$\displaystyle \text{Given: }\,x+\dfrac{1}{x}\,\text{ is a prime number.}$

$\displaystyle \text{Can }\:x^3+\dfrac{1}{x^3}\,\text{ also be a prime number?}$

The only case is: .$\displaystyle x = 1.$

$\displaystyle \begin{array}{cccccccc} \text{Suppose } x\ne1: & \qquad\qquad x + \dfrac{1}{x} &=& p \\ \\[-2mm] \text{Cube both sides:} & \qquad\qquad \left(x + \dfrac{1}{x}\right)^3 &=& p^3 \\ \\[-2mm] & x^3 + 3x + \dfrac{3}{x} + \dfrac{1}{x^3} &=& p^3 \\ \\[-2mm] & \left(x^3 + \dfrac{1}{x^3}\right) + 3\underbrace{\left(x + \dfrac{1}{x}\right)}_{\text{This is }p} &=& p^3 \end{array}$

. . . . . . . . . . . . . . . . . . . .$\displaystyle \begin{array}{cccccc} \left(x^3 + \dfrac{1}{x^3}\right) + 3p &=& p^3 \\ \\[-3mm] \qquad x^3 + \dfrac{1}{x^3} &=& p^3 - 3p \\ \\[-3mm] \qquad x^3 + \dfrac{1}{x^3} &=& \underbrace{p(p^2-3)}_{\text{composite}} \end{array}$

• Aug 6th 2010, 07:52 PM
Also sprach Zarathustra
Soroban you have already 2 points.
• Aug 6th 2010, 09:08 PM
melese
Quote:

Originally Posted by Bruno J.
Let $\displaystyle a=x^3+\frac{1}{x^3}$ and $\displaystyle p=x+\frac{1}{x}$.

We have $\displaystyle p^3 = a+3p$, hence $\displaystyle p | a$, hence no! (Wink)

... and also $\displaystyle a>p$.
• Aug 6th 2010, 09:51 PM
Bruno J.
Quote:

Originally Posted by melese
... and also $\displaystyle a>p$.

Yeah I forgot that part, as others pointed out! :)