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
 A: If $x$ is odd, then $x\equiv1$ or $3\pmod4$; in the former case $x^3+x\equiv2\pmod4$, and in the latter case $x^3+x\equiv30\equiv2\pmod4.$
A: You didn't do your math right Let $x= 2k + 1$ so $x(x^2 + 1) = (2k+1)((2k+1)^2 + 1)=$
$(2k+1)((2k+1)^2 + 1)  = 8k^3 + 12k^2 + 8k +2$ so you had a typo.
But it's easier to do
$(2k+1)^3 + (2k + 1)=$
$(2k)^3 + 3(2k)^2 + 3(2k)+ 1 + (2k + 1)=$
$8k^3 + 12k^2 + 6k + 1 + 2k + 1=$
$8k^3 + 12k^2 + 8k + 2$.
So $4|8, 12, 8$ the $4|8k^3 + 12k^2 + 8k$ and the remainder of $8k^3 + 12k^2 + 8k + 2$ is $2$.
....
Also ... this may be overly abstract:  If we let $x = 2k + 1 = m + 1$ where $m=2k$ is an even number, then $x^3 +x = (m+1)^3 + (m+1) = a_3m^3 + a_2m^2 + a_1m + a_0$, a polynomial.  For $j\ge 2; m^j = (2k)^j = 2^j*k^j=4*2^{j-2}{k^j}$ we have $4|a_jm^j$. 
So the remainder when divided by $4$ will be the same as the remainder of $a_1m + a_0$. 
By binomial theorem:  $(m+1)^3 = \sum_{i=0}^3 {3\choose i}m^i$ so for $(m+1)^3 + (m+ 1)$ we have $a_1 = {3\choose 1} + 1$ and $a_0 = {3\choose 0} + 1$.
${3\choose 1} = 3$ and $3+1=4$ so $4|a_jm$ and the remainder when divided by $4$ will be the same as the remainder of $a_0 = {3\choose 0} + 1$ when divided by $4$.
${3\choose 0} = 1$ and 1+1 = 2
A: $x = 2k + 1; \tag 1$
$x ^3 = 8k^3 + 12k^2 + 6k + 1; \tag 2$
$x ^3 + x =  8k^3 + 12k^2 + 6k + 1 + (2k + 1)$
$ = 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 $2$.
A: 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.
A: Consider $f(x) =x^3+x-2$ instead. This can be factorised as $(x-1)(x^2+x+2)$ instead using the remainder theorem ($1^3+1-2 = 0$).
If $x = 1 \pmod 2$, then $x -1 \equiv 0 \pmod 2$ and $x^2+x+2 \equiv 1^2+1+2 \equiv 4 \equiv 0 \pmod 2$. Therefore  $x^3+x-2 \equiv 0 \pmod 4$, so $x^3+x \equiv 2 \pmod 4$.
