<s>[INST] 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?
 [/INST] $a^3b-ab^3 = ab(a^2-b^2)$. 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 
$
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