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

Let $x=2n+1, then
f(x)=x^3+x=(2n+1)^3+(2n+1)=2+8n^3+12n^2+8n \implies \frac{f(x)}{4}=\frac{2}{4}+2n^3+3n^2+2n.$ So the remaner is 2.
