$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
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:
Bacterius:
your answer : n=0,1,2,7
but here n is a positive integer number so n=0 should be deleted
Albert
I believe some definitions of $\mathbb{N}$ include zero, but most do not.
$\text{here is my solution: for }\,n^3+2\,\mod\,2n+1=0$
$\text{we have }\,8n^3+16\,\mod\,2n+1=0$
$\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.
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.