How i should break up cases (Division in to cases) Show that for every integer $m>3$, at least one of $m, m+2$, or $m+4$ is composite.
I have to do this using division in to cases. Does anyone have any input on how I should break up the cases because I am not too sure.
 A: Either $m$ is divisible by 3, or $m+1$ is divisible by 3, or $m+2$ is divisible by 3. (do you agree with this?)
Note that $m+4=(m+1)+3$. So in case $m+1$ is divisible by 3, then also $m+4$ is divisible by 3.
So indeed, in every case: either $m$, $m+2$ or $m+4$ is divisible by 3 and thus composite.
(Edit: I started with the cases that $m$ is even and that $m$ is odd. If $m$ is even, then it is composite, but this trail wasn't needed in the end.)
A: Hint:
$$m+4 \equiv m+1 \pmod{3}$$
Edit after OP solved the problem:
If $m\equiv 0\pmod{3}$,  $m$ is divisible by $3$  and since $m>3$, it is composite.
If $m\equiv 1 \pmod{3}$, $m+2 \equiv 3\equiv 0 \pmod 3$, and since $m+2>3$, it is composite.
If $m \equiv 2 \pmod{3}$, $m+4 \equiv m +1 \equiv 3 \equiv  0 \pmod 3$, and since $m+4>3$, it is composite.
A: all integers $\mathbb{Z}$ can be classified as members of one of the following classes:
$$C_0=3k$$
$$C_1=3k+1$$
$$C_2=3k+2$$
Such that $\ k\in \mathbb{Z}$.     
Let $m=3k+i \quad \Big| \quad i=0 \lor 1 \lor 2$, then:
$$\begin{align}
m &= 3k+i  && \in C_i \\ 
m+2 &=3k+i+2 && \in C_{i+2 \pmod 3} \\  
m+4 &=3k+i+4=3(k+1)+i+1 &&\in C_{i+1 \pmod 3}
\end{align}$$
So for any $m>3$, you can be sure to find a composite from this set of three. There will be one that belongs in the $C_0$ class.
A: $$m(m+k)(m+2k)\equiv m(m^2-k^2)\pmod3$$
If $3\mid m,3\mid m(m+k)(m+2k)$
else $m\equiv\pm1\implies m^2\equiv1\pmod3$
Similarly if $3\nmid k,k\equiv\pm1\implies k^2\equiv1\pmod3$
$\implies3\mid m(m+k)(m+2k)$ if $3\nmid k\iff(k,3)=1$
Now if $m>3,k\ge0;$ $$m(m+k)(m+2k)>3$$ and is divisible by $3$ hence composite if $3\nmid k\iff(k,3)=1$ 
