Proving that $ 30 \mid ab(a^2+b^2)(a^2-b^2)$ How can I prove that $30 \mid ab(a^2+b^2)(a^2-b^2)$ without using $a,b$ congruent modulo $5$ and then 
$a,b$ congruent modulo $6$ (for example) to show respectively that $5 \mid ab(a^2+b^2)(a^2-b^2)$ and 
$6 \mid ab(a^2+b^2)(a^2-b^2)$? 
Indeed this method implies studying numerous congruences and is quite long.

The intention of this answer is to show you that trying all possibilities is in fact not that long. (Of course, it si more elegant, when you find a solution which avoids trying all possibilities.) Let me start by plotting 5x5 table with all possibilites for the remainders of a and b. (I do not know of a good way of making tables here - I tried something anyway.)
$$
\begin{array}{c|ccccc}
b \backslash a & 0 & 1 & 2 & 3 & 4 \\
\hline
0 &   &   &   &   &   \\
1 &   &   &   &   &   \\
2 &   &   &   &   &   \\
3 &   &   &   &   &   \\
4 &   &   &   &   &   \\
\end{array}
$$
If we rewrite our expression as $ab(a-b)(a+b)(a^2+b^2)$, we see that all possibilities where $a=0$ or $b=0$ are ok (marked by $\circ$). 
$$
\begin{array}{c|ccccc}
b \backslash a & 0 & 1 & 2 & 3 & 4 \\
\hline
0 & \circ & \circ & \circ & \circ & \circ \\
1 & \circ &   &   &   &   \\
2 & \circ &   &   &   &   \\
3 & \circ &   &   &   &   \\
4 & \circ &   &   &   &   \\
\end{array}
$$
Also possibilities where a=b are ok (since $a-b\equiv 0\pmod 5$ and so are those where $a=5-b$ (since $a+b\equiv 0\pmod 5$), hence we can omit both diagonals (marked by $\bullet$).
$$
\begin{array}{c|ccccc}
b \backslash a & 0 & 1 & 2 & 3 & 4 \\
\hline
0 & \circ & \circ & \circ & \circ & \circ \\
1 & \circ & \bullet &   &   & \bullet \\
2 & \circ &   & \bullet & \bullet &   \\
3 & \circ &   & \bullet & \bullet &   \\
4 & \circ & \bullet &   &   & \bullet \\
\end{array}
$$
There are only 8 possibilities left, and since the roles of a and b are symmetric, we only have to try: (1,2), (1,3), (4,2), (4,3). In all these cases $a^2+b^2\equiv 0 \pmod 5$.
