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    #1
    $n\in N $

    $\text{and }(n^3 +2)\,\mod\,(2n+1)=0$

    $\text{please find }n$
    Last edited by MarkFL; January 29th, 2013 at 17:11. Reason: tidy up LaTeX a bit

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    #2
    Quote Originally Posted by Albert View Post
    $ n\in N $

    $ and \,\, (n^3 +2) \,\, mod \,\, (2n+1)=0 $

    $please \,\, find \,\, n $
    Well n = 1 by inspection.

    -Dan

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    #3
    Let $2n + 1 = p$. Then:

    $$n^3 + 2 \equiv \left [ 2^{-1} (p - 1) \right ]^3 + 2 \equiv 2 - 2^{-3} \pmod{p}$$

    Thus, to satisfy your condition, we require:

    $$ 2^{-3} \equiv 2 \pmod{p} ~ ~ \implies ~ ~ 2^{-4} \equiv 1 \pmod{p} ~ ~ \implies 2^4 \equiv 16 \equiv 1 \pmod{p}$$

    Therefore the only solutions for $p$ are $\{1, 3, 5, 15\}$, which translate to $n = \{0, 1, 2, 7\}$.

    May be verified experimentally:


  4. MHB Master
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    #4 Thread Author
    Bacterius:
    your answer : n=0,1,2,7
    but here n is a positive integer number so n=0 should be deleted

    Albert

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    #5
    I believe some definitions of $\mathbb{N}$ include zero, but most do not.

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    #6 Thread Author
    $\text{here is my solution: for }\,n^3+2\,\mod\,2n+1=0$
    $\text{we have }\,8n^3+16\,\mod\,2n+1=0$
    $\text{but }\,8n^3+16\,\mod\,2n+1=15 $
    $\therefore\,15\,\mod\,2n+1=0$
    $\text{and }\,n=\begin{Bmatrix}
    1,2,7
    \end{Bmatrix}$
    Last edited by Albert; January 29th, 2013 at 23:28.

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    #7
    Albert,

    Just a LaTeX tip...to include text within your code, so that it is not italicized like the variables are, use the \text{} command, e.g., \text{insert text here}. Leave everything else outside of the braces.

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    #8
    Quote Originally Posted by Albert View Post
    Bacterius:
    your answer : n=0,1,2,7
    but here n is a positive integer number so n=0 should be deleted

    Albert
    Your definition of $\mathbb{N}$ was unclear, thus I conservatively included $n = 0$.

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