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.
 A: 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 $2$)
A: You need to show $ab(a^2 - b^2)(a^2 + b^2)$ is a multiple of 2,3, and 5 for all $a$ and $b$.
For 2: If neither $a$ nor $b$ are even, they are both odd and $a^2 \equiv b^2 \equiv 1 \pmod 2$, so that 2 divides $a^2 - b^2$. 
For 3: If neither $a$ nor $b$ are a multiple of 3, then $a^2 \equiv b^2 \equiv 1 \pmod 3$, so 3 divides $a^2 - b^2$ similar to above. 
For 5: If neither $a$ nor $b$ are a multiple of 5, then either $a^2 \equiv 1 \pmod 5$ or $a^2 \equiv -1 \pmod 5$. The same holds for $b$. If $a^2 \equiv b^2 \pmod 5$ then 5 divides $a^2 - b^2$, while if $a^2 \equiv -b^2 \pmod 5$ then 5 divides $a^2 + b^2$.
This does break into cases, but as you can see it's not too bad to do it systematically like this.
A: 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 $a^2+b^2\equiv 0 \pmod 5$.
A: Below I show it is a very special case of  following generalization of the Euler-Fermat theorem.
Theorem $\  $ Suppose that $\rm\ n\in \mathbb N\ $ has the prime factorization $\rm\:n = p_1^{e_{\:1}}\cdots\:p_k^{e_k}\ $ and suppose  that for all $\rm\,i,\,$  $\rm\ e\ge e_i\ $ and $\rm\ \phi(p_i^{e_{\:i}})\mid f.\ $ Then $\rm\ n\mid (ab)^e\,(a^f-b^f)\ $ for all $\rm\: a,b\in \mathbb Z.$ 
Proof $\ $ Notice that if $\rm\ p_i\mid ab\ $ then $\rm\:p_i^{e_{\:i}}\ |\ (ab)^e\ $ by $\rm\ e_i \le e.\: $ Else $\rm\:a\:$ and $\rm\:b\:$ are coprime to $\rm\: p_i\:$ so by Euler's phi theorem, $\rm\ mod\ q = p_i^{e_{\:i}}\!: \ a^{\phi(q)}\equiv 1\equiv b^{\phi(q)} \Rightarrow\ a^f\equiv 1\equiv b^f\: $ by $\rm\: \phi(q)\mid f.\ $  Thus since all $\rm\ p_i^{e_{\:i}}\ |\  (ab)^e\ (a^f - b^f\:)\ $ so too does their lcm = product = $\rm\: n.$
Corollary $\rm\ \ 6p\mid a\: b^p - b\: a^p\ $ for all $\rm\:a,b\in \mathbb Z,\:$ prime $\rm\:p\ne 2,3$ 
Specializing $\rm\ p = 5\ $ yields the result sought in the question.
Remark $\: $ Note that the proof of the general theorem is less work than some other direct proofs of your special case. For some other variations (e.g. Carmichael's) see my posts here.
A: Hint for the divisors 2 and 3.
Note the factorisation $ab(a+b)(a-b)(a^2+b^2)$ which is divisible by $ab(a-b)(a+b)$
Either one of $a,b$ is divisible by 2 (or 3) or not. If not, look at the other factors to complete the proof.
Studying cases does not look too onerous to me.
