Prove that 360 divides (a-2)(a-1)a.a.(a+1)(a+2)

Question

there's a question which asks to prove that

360 | a²(a²-1)(a²-4)

I attempted it in the following manner.

a²(a²-1)(a²-4) = (a-2)(a-1)(a)(a+1)(a+2)(a)

The first 5 terms represent the product of 5 consecutive terms. Hence, One of them will have a factor of 5 One of them will have a factor of 4 One of them will have a factor of 3 One of them will have a factor of 2 => 5 x 4 x 3 x 2 = 120 will divide the given expression for sure.

Now, how do i bring a factor of 3. I know that i have missed one power of a. But, i think i am missing something. Help will be appreciated

Thank you

Answer 1

Hint: both $(a-2)(a-1)a$ and $a(a+1)(a+2)$ are products of three consecutive integers.

Answer 2

We know from here or here, the product $n$ consecutive integers is divisible by $n!$ for integer $n>0$

$$(a-2)(a-1)a\cdot a(a+1)(a+2)$$

$$=(a-2)(a-1)a(a+3-3)(a+1)(a+2)$$

$$=\underbrace{(a-2)(a-1)a(a+1)(a+2)(a+3)}_{\text{ the product of } 6\text{ consecutive integers }}-3\underbrace{(a-2)(a-1)a(a+1)(a+2)}_{\text{ the product of } 5\text{ consecutive integers }}$$

We could divide $a$ as $a-3+3$ as well.

Answer 3

If 3 divides $a-1$ then it must also divide $a+2$. Similarly, if it divides $a-2$ it must divide $a+1$. If it divides neither of these, then it divides $a$. In all three cases you get a factor of nine to go along with your known factors of $5$, $4$, and $2$.

Answer 4

Your existing argument shows that the product is divisible by $8$ and $5$, and hence by $40$. To show that it is divisible by $9$, note that either $3$ divides $a$, in which case it occurs twice because of $a^2$, or it divides two of the other factors.

Answer 5

We have $360 = 8 \times 9 \times 5$, so it will suffice to check that $8,9,5|a^2(a^2-1)(a^2-4)$.

For $9$, note that $a^2$ is congruent to either $1$ or $0$ modulo $3$, so both $a(a^2-1)$ and $a(a^2-4)$ are divisible by $3$, and consequently $9|a^2(a^2-1)(a^2-4)$.

Likewise, $a^2$ is congruent to either $1$ or $0$ or $4$ modulo $5$, so again the divisibility $5|a^2(a^2-1)(a^2-4)$ holds.

Finally, we look at divisibility by $8$. If $a$ is even, then $4|a^2$ and $4|a^2 -4$, so you have the divisibility $8|a^2(a^2-1)(a^2-4)$ (even with $16$ in place of $8$). If $a$ is odd, then $a-1,a+1$ are even and one of them is divisible by $4$, so $a^2-1$ is divisible by $8$ and the sought divisibility holds.

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As a general rule, if you have a problem like (i.e. "Show that $360|f(a)$ for all $a$") this and don't want to think too much, you can first writhe $360$ as a product of powers of primes (nothing special about $360$ here, of course), and then check that if $P$ is a highest power of a prime dividing $360$ (i.e. $P=8,9,5$ here) you always have $P|f(a)$. This second check can be done by brute force: just check that $P|f(0),f(1),\dots,f(P-1)$ - this requires just the amount of computation that can be done by hand.

Answer 6

$$ \begin{align} (a-2)(a-1)a\cdot a(a+1)(a+2) &=5!\,a\binom{a+2}{5}\\ &=120[(a+3)-3]\binom{a+2}{5}\\ &=120(a+3)\binom{a+2}{5}-360\binom{a+2}{5}\\ &=120\cdot6\binom{a+3}{6}-360\binom{a+2}{5}\\ &=360\left[2\binom{a+3}{6}-\binom{a+2}{5}\right] \end{align} $$

Answer 7

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