Thread: Proove that...

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

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

Originally Posted by Sorombo

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

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

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

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

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

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

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

or:

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

Now consider the left hand side modulo 3.

CB