Proove that...

**Sorombo** · September 19th 2011, 02:24 PM

Proove that for every $n \in \mathbb{N}$

$n^3-9n+27\not\equiv 0 \mod 81$

**CaptainBlack** · September 21st 2011, 12:15 AM

Originally Posted by Sorombo

Proove that for every $n \in \mathbb{N}$

$n^3-9n+27\not\equiv 0 \mod 81$

Suppose otherwise, then there exists a $k\in \mathbb{Z}$ such that:

$n^3-9n+27=k\times 81=k\times 9^2$

which implies that $9|n^3$ which in turn implies that $3|n$ .

Hence there exists a $\lambda \in \mathbb{N}$ such that:

$27 \lambda^3-27 \lambda +27=3 \times k \times 27$

or:

$\lambda^3- \lambda +1=3 \times k$

Now consider the left hand side modulo 3.

CB

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