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  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Primes...

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

    Can x^3+\frac{1}{x^3} also be a prime number?
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    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!
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    One point to Bruno J.
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    except the case  x = 1


     p^2 - 3  may equal  1
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    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}<br />
\text{Suppose } x\ne1: & \qquad\qquad x + \dfrac{1}{x} &=& p \\ \\[-2mm]<br />
\text{Cube both sides:} & \qquad\qquad \left(x + \dfrac{1}{x}\right)^3 &=& p^3 \\ \\[-2mm]<br /> <br />
& x^3 + 3x + \dfrac{3}{x} + \dfrac{1}{x^3} &=& p^3 \\ \\[-2mm]<br /> <br />
& \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}<br />
\left(x^3 + \dfrac{1}{x^3}\right) + 3p &=& p^3 \\ \\[-3mm]<br />
\qquad x^3 + \dfrac{1}{x^3} &=& p^3 - 3p \\ \\[-3mm]<br />
\qquad x^3 + \dfrac{1}{x^3} &=& \underbrace{p(p^2-3)}_{\text{composite}}<br />
\end{array}

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  6. #6
    MHF Contributor Also sprach Zarathustra's Avatar
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    Soroban you have already 2 points.
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    Quote Originally Posted by Bruno J. View Post
    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 .
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  8. #8
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by melese View Post
    ... and also a>p .
    Yeah I forgot that part, as others pointed out!
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