Find all $x$ such that $x^{35} + 5x^{19} + 11x^3$ is divisible by 17. 
Find all $x$ such that $x^{35} + 5x^{19} + 11x^3$ is divisible by 17.

I think we can use the fact that we can mod everything by 17 and want 0. But how exactly should we go about doing this?
 A: \begin{align}x^{35}+5x^{19}+11x^3
&= x^{17}\cdot x^{17} \cdot x + 5 x^{17}\cdot x^2 + 11 x^3\\
&\equiv x\cdot x \cdot x + 5x \cdot x^2 + 11 x^3 \;(\mathrm{mod}\;17)
= 17x^3\\
&\equiv 0\;(\mathrm{mod}\; 17)
\end{align}
where we apply Fermat's little theorem a few times ($a^p \equiv a \;(\mathrm{mod}\; p)$). Therefore it works for all integers $x$.
A: $$x^{17}+5x^{19}+11x^3=(x^{18}+6x^2)(x^{17}-x)+17x^3$$
Since $x^{17}-x$ is divisible by $17$ and so is $17x^3$, so is the original polynomial.
A: Any integer value for $x$ will do !
Observe that 
\begin{eqnarray*}
x^{16} \equiv 1 \pmod{17} 
\end{eqnarray*}
and the expression can be written as $x^3(11+5x^{16}+x^{32})$ so the bracket will always be divisible by $17$.
A: Lil' Fermat says this is the same as solving the equation in the field $\mathbf Z/17\mathbf Z$
$$x\cdot (x^{17})^2+5x^{17}\cdot x^2+11x^3+=(1+5+11)x^3=0.$$
Thus the given expression is divisible by $17$  for all  $x$.
A: $\begin{align}{\rm mod}\ 17\!:\,\ x\not\equiv 0\,\Rightarrow\, x^{\large 16}\!\equiv 1\,\Rightarrow\, &\ x^{\large\color{#c00}{n}}\equiv\, x^{\large\color{#c00}{ n\bmod 16}}\ \ \text{by little Fermat, hence}\\[.3em]
&\ x^{\large\color{#c00}{35}}\!+5 x^{\large\color{#0a0}{19}}\!+ 11 x^{\large 3}\\[.2em]
\equiv\ &\  x^{\large\color{#c00}3}\ + 5 x^{\large\color{#0a0}3} +\, 11 x^{\large 3}\ \ \,{\rm by}\,\ \ {\rm mod}\ 16\!:\ \color{#c00}{35\equiv 3},\ \color{#0a0}{19\equiv 3}\\[.2em]
\equiv\  & 17x^{\large 3}\equiv\, 0
\end{align}$
