Prove $30|(a^3b-ab^3) $ Prove that if three distinct integers are chosen at random then there will exists two among them, say $a$ and $b$ such that $30|(a^3 b-ab^3)$
 A: Hints:
$$a^3b-ab^3=ab(a^2-b^2)=ab(a-b)(a+b)$$
Hints: check that no matter what the parity of $\;a,b\;$ is, the above product is always even and thus divisible by two. 
Also, no matter what $\;a,\,b\,$ are, the product is always divisible by $\;3\;$ .
I'll leave to you the funniest part: to prove the product of two of the three different integers chosen is always divisible by $\;5\;$ (not necessarily a given pair $\;a,b\;$ !) . Observe that you can reduce this to the case where the three different integers are different $\;\pmod5\;$ , otherwise we're done (why?) .
A: By Fermat's little theorem (or simply by direct calculation), $x^2\equiv x$ mod $2$ and $x^3\equiv x$ mod $3$ for any integer $x$.  It follows that
$$a^3b\equiv ab\equiv ab^3\mod 6$$
for any pair of integers $a,b$, hence $6$ divides $a^3b-ab^3$ for any pair.  It remains to show that $5$ divides $a^3b-ab^3=ab(a^2-b^2)$ for at least one of the three pairs.  But there are only three squares mod $5$, namely $0$, $1$, and $-1$.  So given three integers, either two of their squares, say $a^2$ and $b^2$ belong to the same residue class, in which case $5$ divides $a^2-b^2$, or one of the three squares, say $a^2$ is $0$ mod $5$, in which case $5$ divides $a$ (and you can let $b$ be either of the other two numbers).  In either case $5$ divides $ab(a^2-b^2)=a^3b-ab^3$ for the chosen pair, and we are done.
A: 1) Divisibility by $5$.
Let $\{a,b,c\}$ the three numbers reduced modulo $5$. If one of them is $0$ or two of them are equal we clearly have
$$ab(a+b)(a-b)\equiv 0\pmod 5$$
So be all the three, $a,b,c$ distinct and non-zero modulo $5$ so we have
$$\{a,b,c\}=\{1,2,3\},\{1,2,4\},\{1,3,4\},\{2,3,4\}$$
In the four cases we clearly have two of the (class) numbers have sum $0$ modulo $5$ (because of $2+3=1+4$) so $ab(a+b)(a-b)\equiv 0\pmod 5$.
Thus the expression is for two of the three integers divisible by $5$.
2) Divisibility by $3$ and by $2$.
The same method but easier than for divisibility by $5$ can be used for divisibility by $3$ and by $2$.
Thus the divisibility by $30=2\cdot3\cdot5$ of the expression for two of the three numbers..
