# Primes...

• Aug 6th 2010, 02:25 PM
Also sprach Zarathustra
Primes...
Given that $x+\frac{1}{x}$ is a prime number.

Can $x^3+\frac{1}{x^3}$ also be a prime number?
• Aug 6th 2010, 03:57 PM
Bruno J.
Quote:

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! (Wink)
• Aug 6th 2010, 04:19 PM
Also sprach Zarathustra
One point to Bruno J.
• Aug 6th 2010, 06:58 PM
simplependulum
except the case $x = 1$ (Happy)

$p^2 - 3$ may equal $1$
• Aug 6th 2010, 08:34 PM
Soroban
Hello, Also sprach Zarathustra!

Quote:

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

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

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}$

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

• Aug 6th 2010, 08:52 PM
Also sprach Zarathustra
Soroban you have already 2 points.
• Aug 6th 2010, 10:08 PM
melese
Quote:

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! (Wink)

... and also $a>p$.
• Aug 6th 2010, 10:51 PM
Bruno J.
Quote:

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

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