Confirmation in one step involved in solving 'Show that any 2 of 3 integers chosen randomly satisfy $10|a^3b - ab^3$.

Question

After factoring: $ab(a-b)(a+b)$ gives that it is always multiple of 2 so I only need to prove it is multiple of 5. For that I made a set $A={a,b,c,(a-b),(a+b),(b-c),(b+c),(c-a),(c+a)}$
By pigeon hole principle, difference of any 2 elements must give a multiple of 5 and absolute difference of any two elements in A should also be an absolute of one element in A. Thus $10|a^3b - ab^3$

Is the method used to prove for multiple of 5 acceptable?

Answer 1

HINT:

$$ab(a^2-b^2)$$

Now $x^2 \in \{ \pm 1\} \mod 5$, so two of the squares will be congruent $\mod 5$.

Answer 2

Suppose $5$ does not divide any of $a,b,c$.

Working modulo $5$,we can let $\{a,b,c\}\subset \{1,2,3,4\}.$

Suppose $5\not |\,(a-b)(a+b).$ Then $a\ne b$ and $a\ne 5-b$ and $b\ne 5-a$. It is not possible that $b=5-b$ or $a=5-a$. And $5-b=5-a$ implies $a=b.$ So $a,b, 5-b, 5-a$ are $4$ distinct values.

So $\{a,b,5-b,5-a\}= \{1,2,3,4\}.$

Now $c\in \{1,2,3,4\}.$ So $(\,c=a \lor c=b \lor c=5-b \lor c=5-a\,),$ implying $(\,5|c-a \lor 5|c-b \lor 5|c+b \lor 5|c+a\,).$