Why is $2x^3 + x$, where $x \in \mathbb{N}$, always divisible by 3? So, do you guys have any ideas? Sorry if this might seem like dumb question, but I have asked everyone I know and we haven't got a clue.
 A: certainly true for $n = 1$. IF $2n^3 + n \cong_3 0 $, then
$$ 2(n + 1)^3 + n + 1 = 2(n^3 + 3n^2 + 3n + 1) + n + 1 = 2n^3 +6n^2 + 6n +2 + n + 1 $$
$$ = (2n^3 + n ) + 3(2n^2 + 2n + 1) \cong_3 0$$
REsult follows from math induction.
A: “Fermat's little theorem” says that, if $p$ is a prime, then $a^p\equiv a\pmod{p}$, for all integers $a$.
Then
$$
2x^3+x\equiv2x+x\equiv3x\equiv0\pmod{3}
$$
A: $2x^3+x=x\cdot(2x^2+1)$. If $3|x$, then the result follows immediately. Otherwise, $x=3k\pm1\rightarrow$ $\iff2(3k\pm1)^2+1=2(9k^2\pm6k+1)+1=18k^2\pm12k+3=3(6k^2\pm4k+1)$. QED.
A: $2x^3+x=3x^3-x(x-1)(x+1)$; since $x-1,x,x+1$ are three consecutive integers, one of them is divisible by $3$.
A: $x \mod 3=\{0,1,2\}$ for $x\in \mathbb{N}$. 
In the other words, if you divide any natural number by $3$, the remainder will be $0$ or $1$ or $2$.


*

*$x \mod 3=0 \Rightarrow (2x^3 + x) \mod 3 = 0$

*$x \mod 3=1 \Rightarrow (2x^3 + x) \mod 3 = 0$

*$x \mod 3=2 \Rightarrow (2x^3 + x) \mod 3 = 0$

A: 2x3 + x = x(2x2 + 1) = x(2x2 - 2 + 3) = x[2(x + 1)(x - 1) + 3] = 3x + 2x(x + 1)(x - 1)
3x is always divisible by 3 and one of x - 1, x, x + 1 is also always divisble by 3
A: This should be clear: $2x^3+x\equiv x-x^3(\mod 3)\equiv -x(x-1)(x+1)(\mod 3)\equiv 0$.
A: By the Fermat's little theorem,
$$x^3 \equiv x \pmod 3$$
, therefore
$$2x^3 + x \equiv 2x + x \equiv 3x \equiv 0 \pmod 3$$
