Prove this Equation is Divisible By... Prove that among any three distinct integers we can find two, say $a$ and
$b$, such that the number $a^3b − ab^3$ is divisible by $10$.
Can anyone help me solve this?
 A: Write
$$a^3b - ab^3 = ab(a^2 - b^2) = ab(a - b)(a + b)$$
and show for some choice of $a,b$ among any three distinct integers, the expression on the right is divisible by $2$ and $5$ (hint: divisibility by $2$ is automatic)
A: $a^3b-ab^3 = ab(a^2-b^2)$. If both $a$ and $b$ are odd, $(a^2-b^2)$ is even, so the expression is guaranteed to be even.
Now consider three arbitrary integers $\{c,d,e\}$. If any of these is divisible by $5$, we can chose this and either of the other two to make the target expression divisible by $5$. Otherwise we have $c^2 \equiv \{-1,1\} \bmod 5$, and similarly for $d$ and $e$. Since there are only two choices for the value $\bmod 5$, at least two of $\{c,d,e\}$ must have the same value $\bmod 5$ when squared. Then, choosing these two as $a$ and $b$, we see that $(a^2-b^2)\equiv 0 \bmod 5$, so $5$ divides $(a^2-b^2)$, and thus the target expression $a^3b-ab^3$ is divisible by $10$.

$ (5k+1)^2 = 25k^2+10k + 1  = 5(5k^2+2k) + 1 \\
 (5k+2)^2 = 25k^2+20k + 4  = 5(5k^2+4k + 1) - 1 \\
 (5k+3)^2 = 25k^2+30k + 9  = 5(5k^2+6k + 2) - 1 \\
 (5k+4)^2 = 25k^2+40k + 16  = 5(5k^2+8k + 3) + 1 
$
A: To prove atleast one of either $(a-b)$ or $(a+b)$ is divisible by $10$:  

$$a\equiv0,\pm1,\pm2,\pm3,\pm4,5\pmod{10}\implies a^2\equiv0,1,4,9,6,5$$ So, there are $6$ in-congruent squares $\pmod{10}$. So, if we choose more than $6$ distinct integers, at least two squares will be congruent $\pmod{10}$ .

We then have $a^3b-ab^3= ab(a-b)(a+b)$, so our theorem is proved.
