Prove this Equation is Divisible By...

Question

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?

Answer 1

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)

Answer 2

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.

Answer 3

$a^3b-ab^3 = ab(a^2-b^2)$. 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 $10$.

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$ (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 $