<s>[INST] 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.
 [/INST] 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 $a^2+b^2\equiv 0 \pmod 5$.
</s>