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

5 Answers

Relevance
  • 1 decade ago
    Favourite answer

    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.

    • Commenter avatarLog in to reply to the answers
  • 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

    • Commenter avatarLog in to reply to the answers
  • 1 decade ago

    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.

    • Commenter avatarLog in to reply to the answers
  • 1 decade ago

    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.

    • Commenter avatarLog in to reply to the answers
  • What do you think of the answers? You can sign in to give your opinion on the answer.
  • 1 decade ago

    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

    • Commenter avatarLog in to reply to the answers
Still have questions? Get answers by asking now.