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.

Hint: First write $f(a,b)=ab(a^2+b^2)(a^2-b^2)=a^5b-b^5a$.  Recall Fermat's Little Theorem which states $x^{p-1}\equiv 1 \pmod{p}$ for $x$ not divisible by $p$.
Assume $5$ does not divide either of $a,b$. (otherwise it follows automatically that $5|f(a,b)$)  Then
$$a^5b-b^5a\equiv ab-ba\equiv 0\pmod{5}$$ and we see that $5|f(a,b)$.
You can prove that $2$ and $3$ divide $f(a,b)$ in the same way. 
(An alternative way to see that it is divisible by $2$ is to note that one of $a$, $b$, $a^2+b^2$ must be divisible by $2$)
