If three distinct integers are chosen at random, show that there will exist two among them, say $a$ and $b$, such that $30 | (a^3b-ab^3)$ Problem:

If three distinct integers are chosen at random, show that there will exist two among them, say $a$ and $b$, such that $30 | (a^3b-ab^3)$

My work:
$a^3b-ab^3=ab(a+b) (a-b)$ and if $30 | (a^3b-ab^3)$, then each of $2,3,5$ divides $(a^3b-ab^3)$ since $2\cdot3\cdot5=30$ and $(2,3,5)=1$
Case for 2:
If 2 divides either $a$ or $b$, then all good.
If not, then $a$ and $b$ are both odd and so $a+b$=sum of odds = even is divisible by 2. Done.
Case for 3:
If 3 divides a or b, good.
Else, let:

*

*$a=3k+1$ and $b=3k+1$. Then, $3|(a-b)$

*$a=3k+1$ and $b=3k+2$. Then, $3|(a+b)$

*$a=3k+2$ and $b=3k+2$. Then, $3|(a-b)$

*$a=3k+2$ and $b=3k+1$. Then, $3|(a+b)$
So far, so good.
Case for 5:
Completely stuck here. Five possible remainders ($0,1,2,3,4$) seem to be impossible to do. Any hints?
Questions:

*

*Need hints for last case.


*Why do we need to pick three integers when we only need to check with two?

You can remove "distinct" to make it slightly stronger (but it's a trivial case). Notice $30=2\cdot 3\cdot 5$ and $2,3,5$ are primes. Your problem is: Among any three integers $a,b,c$, exist at least two $x,y\in\{a,b,c\}$ such that $2,3,5\mid xy(x+y)(x-y)$.
By Pigeonhole principle, either $abc\equiv 0\pmod{5}$ or $a\equiv \pm b\pmod{5}$ or $a\equiv \pm c\pmod{5}$ (because $a,b,c$ are three integers that belong to $0,1,2,-2,-1$ mod $5$), so exist $x,y\in\{a,b,c\}$ such that $5\mid xy(x+y)(x-y)$.
But then by Pigeonhole principle either $xy\equiv 0\pmod{3}$ or $x\equiv \pm y\pmod{3}$. Similarly, either $xy\equiv 0\pmod{2}$ or $x\equiv \pm y\pmod{2}$.
One generalization can be: Among any $k\ge 3$ integers $a_1,a_2,\ldots, a_k$ exist $x,y\in\{a_1,a_2,\ldots,a_k\}$ such that $6h\mid xy(x+y)(x-y)$, where $h\ge 5$ is any integer such that $\gcd(h,6)=1, \frac{h-1}{2}<k$. 
