Question

for any positive integer n prove that n³_n is divisible by 6​

Answer 1

Answer:

Let us consider

a = n³ – n

a = n (n² – 1)

a = n (n + 1) (n – 1)

Assumtions:

1. Out of three (n – 1), n, (n + 1) one must be even, so a is divisible by 2.

2. (n – 1) , n, (n + 1) are consecutive integers thus as proved a must be divisible by 3.

From (1) and (2) a must be divisible by 2 × 3 = 6

Thus, n³ – n is divisible by 6 for any positive integer n.

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Answer 2

So we have to prove,

${n}^{3} - n\equiv0 \: \: ( \rm{mod} \: 6)\:\: \forall\:\: n \in \mathbb{Z}^+.$

$\longrightarrow6 \: | \: {n}^{3} - n$

But before this we must prove another conclusion.

We have to prove,

"Product of any three consecutive integers is always divisible by 3."

Let the integers be $n-1,\: n$ and $n+1$

On dividing then by 3, the possible remainders would be 0, 1 and 2.

So the three consecutive integers would be in the form of $3k,\ 3k + 1$ or $3k + 2\:\: \forall \:\:k \in \mathbb{Z}.$

Case 1:

If $n = 3k$

As $3\; |\; 3k$

$\longrightarrow 3 \;|\; n$

$\longrightarrow 3\; |\; (n - 1)n(n + 1)$

Case 2:

If $n = 3k + 1$

Then,

$\longrightarrow n - 1 = 3k$

As $3\; | \;3k$

$\longrightarrow 3\; |\; n - 1$

$\longrightarrow3\; | \;(n - 1)n(n + 1)$

Case 3:

If $n = 3k + 2$

Then,

$\longrightarrow n + 1 = 3k + 3$

$\longrightarrow n + 1 = 3(k + 1)$

As $3 \;|\; 3(k + 1),$

$\longrightarrow 3 \;|\; n + 1$

$\longrightarrow 3 \;|\; (n - 1)n(n + 1)$

Hence, the product of three consecutive integers is always divisible by 3.

Now coming to the required proof,

Given that,

$n^3 - n\:\: \forall\:\:n \in \mathbb{Z}^+.$

$=n(n^2 - 1)$

• Expanding $(n^2-1)$ using a² - b² = (a - b)(a + b),

$=(n-1)n(n+1)$

We can observe that they are three consecutive integers.

From here, we can immediately draw two conclusions.

Conclusion 1:

Since they are consecutive integers, atleast one of them must be even and divisible by 2.

Hence,

$2\; |\; (n - 1)n(n + 1)$

$\longrightarrow 2 \;| \;n^3 - n \quad\quad\dots(1)$

Conclusion 2:

The product of three consecutive integers is divisible by 3 (proved earlier).

Hence, $3\; | \;(n - 1)n(n + 1)$

$\longrightarrow 3 \;|\; n^3 - n\quad\quad\dots (2)$

From (1) and (2),

$\longrightarrow \underline{\underline{6\:\:|\:\:{n}^{3} - n}}$