# prove that for any integer n,3|n^3 2n

Relevance

we can use induction .. . .

n = 1

3 | 1^3 + 2 = 3 True

then

suppose it is true for n = k

then

(k^3 + 2k) = 3m for some integer m

then we have to show it is true for n = k+1

(k+1)^3 + 2(k+1)

= k^3 + 3k^2 + 3k + 1 + 2k + 2

= k^3 + 2k + 3k^2 + 3k + 3

= 3m + 3k^2 + 3k + 3

= 3 (m + k^2 + k + 1)

thus

(k+1)^3 + 2(k+1) is divisible by 3

.. . .. .

3| (n^3 + 2n) for all integers n.

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• I think you mean n^3 + 2n

Induction will solve this.

For base case n = 1, 1^3 + 2 = 3 is divisible by 3.

Assume this is true for all n = k

-> k^3 + 2k is divisible by 3.

Let k^3 + 2k = 3m (m is an integer)

Proving for n = k + 1

(k + 1)^3 + 2(k + 1) = k^3 + 1 + 3k^2 + 3k + 2k + 2

= (k^3 + 2k) + 3k^2 + 3k + 3

= 3m + 3k^2 + 3k + 3

= 3(m + k^2 + k + 3)

which is divisible by 3

Since n = k is true implies n = k + 1 is true, and since the base case is true as well, n^3 + 2n is divisible by 3 for all n ∈ Z

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• n³ + 2n = n(n² + 2)

0 and 1 are the only quadratic residues modulo 3. In other words, if n is not congruent to 0 mod 3, n² is congruent 1 mod 3.

If n ≡ 1 (mod 3), n² + 2 ≡ 0 (mod 3)

If n ≡ 2 (mod 3), n² + 2 ≡ 0 (mod 3)

Hence, if n ≡ 1 (mod 3) or n ≡ 2 (mod 3), 3 | (n³ + 2n).

The remaining case is if n ≡ 0 (mod 3). This obviously means n³ + 2n is divisible by 3.

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• n^3 + 2n

= n^3 - 3n^2 + 3n^2 - 9n + 11n - 33 + 33

= n^2(n - 3) + 3n(n - 3) + 11(n - 3) + 33

= (n - 3)(n^2 + 3n + 11) + 33

=> (n - 3)(n^2 + 3n + 11) is divisible by n = 3

and second term 33 is divisible by 3.

=> n^3 + 2n is divisible by 3.

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• n can only be of three different types of numbers

3m

3m +1

3m +2

If n = 3m, then it's obvious it is going to be divisible by 3

if n = 3m +1, cubing it will be so that there will be a remainder of 1 when divided by 3.

The remainder of 1 plus 2(3m+1)

1 + 6m +2 = 3 +6m will be divisible by 3

If n = 3m +2, cubing it will give a remainder of 2 when divided by 3.

The remainder of 2 plus 2(3m +2)

2 + 6m +4 = 6 +6m will be divisble by 3

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