Primes...

**Also sprach Zarathustra** · August 6th 2010, 01:25 PM

Given that $x+\frac{1}{x}$ is a prime number.

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

**Bruno J.** · August 6th 2010, 02:57 PM

Originally Posted by Also sprach Zarathustra

Given that $x+\frac{1}{x}$ is a prime number.

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

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

We have $p^3 = a+3p$ , hence $p | a$ , hence no!

**Also sprach Zarathustra** · August 6th 2010, 03:19 PM

One point to Bruno J.

**simplependulum** · August 6th 2010, 05:58 PM

except the case $x = 1$

$p^2 - 3$ may equal $1$

**Soroban** · August 6th 2010, 07:34 PM

Hello, Also sprach Zarathustra!

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

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

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

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

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

**Also sprach Zarathustra** · August 6th 2010, 07:52 PM

Soroban you have already 2 points.

**melese** · August 6th 2010, 09:08 PM

Originally Posted by Bruno J.

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

We have $p^3 = a+3p$ , hence $p | a$ , hence no!

... and also $a>p$ .

**Bruno J.** · August 6th 2010, 09:51 PM

Originally Posted by melese

... and also $a>p$ .

Yeah I forgot that part, as others pointed out!

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