if $x$ is odd, show that $x^3+x$ has a remainder 2 when divided by 4 I did part of this question but am stuck and don't know how to continue
I let $x= 2k +1$
Also noticed that $x^3+x = x(x^2+1)$
therefore
$4m+2 = 2k+1((2k+1)^2+1)$
I simplified this and ended up with
$4m+2 = 8k^3+12k^2+8k+2$
I don't know how to continue from and prove that $x^3+x$ has remainder 2 when divided by 4

$x = 2k + 1; \tag 1$
$x ^3 = 8k^3 + 12k^2 + 6k + 1; \tag 2$
$x ^3 + x =  8k^3 + 12k^2 + 6k + 1 + (2k + 1)$
$ = 8k^3 + 12k^2 + 8k +  2 = 4(2k^3 + 3k^2 + 2k)  + 2, \tag 3$
that is,
$x^3 + x \equiv 2 \mod 4, \tag 4$
which by the Euclidean division implies the remainder of $x^3 + x$ when divided by $4$ is $2$.
