Find 'n' in modulo equation

[Albert]

$n\in N $

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

$\text{please find }n$

 

[topsquark]

$ n\in N $

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

$please \,\, find \,\, n $

Well n = 1 by inspection.

-Dan

 

[Bacterius]

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:

>>> for n in range(10000):
...     p = 2 * n + 1
...     r = (pow(n, 3, p) + 2) % p
...     if r == 0:
...             print(n)
... 
0
1
2
7
>>>

 

[Albert]

Bacterius:
your answer : n=0,1,2,7
but here n is a positive integer number so n=0 should be deleted

Albert

 

[MarkFL]

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

 

[Albert]

$\text{here is my solution: for }\,n^3+2\,\mod\,2n+1=0$
$\text{we have }\,8n^3+16\,\mod\,2n+1=0$
$\text{but }\,8n^3+16\,\mod\,2n+1=15 $
$\therefore\,15\,\mod\,2n+1=0$
$\text{and }\,n=\begin{Bmatrix}
1,2,7
\end{Bmatrix}$

 

[MarkFL]

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.

 

[Bacterius]

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 $n = 0$.