Results 1 to 5 of 5

Math Help - Need Help Please (Prime or Not Prime)

  1. #1
    Newbie
    Joined
    Oct 2010
    Posts
    4

    Need Help Please (Prime or Not Prime)

    I need help solving the following question:

    Prove or disprove the following:

    There exists an integer n such that (8n^3 + 12n^2 + 6n + 1) is a prime number.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by mathnb View Post
    I need help solving the following question:

    Prove or disprove the following:

    There exists an integer n such that (8n^3 + 12n^2 + 6n + 1) is a prime number.
    Hint: We can factor this!

    \displaystyle 8n^3 + 12n^2 + 6n + 1 = 8n^3 + 1 + 12n^2 + 6n ........now factor by grouping

    \displaystyle = (2n + 1)(4n^2 - 2n + 1) + 6n(2n + 1) ................the first pair is the sum of two cubes...

    \displaystyle = (2n + 1)(4n^2 + 4n + 1)

    \displaystyle = (2n + 1)^3

    Now what can you say? We can factor this, if we assume it is prime, what must happen? Could it work?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2010
    Posts
    4
    Quote Originally Posted by Jhevon View Post
    Hint: We can factor this!

    \displaystyle 8n^3 + 12n^2 + 6n + 1 = 8n^3 + 1 + 12n^2 + 6n ........now factor by grouping

    \displaystyle = (2n + 1)(4n^2 - 2n + 1) + 6n(2n + 1) ................the first pair is the sum of two cubes...

    \displaystyle = (2n + 1)(4n^2 + 4n + 1)

    \displaystyle = (2n + 1)^3

    Now what can you say? We can factor this, if we assume it is prime, what must happen? Could it work?
    Thanks for the quick reply

    We can disprove the statement, i.e. prove that \displaystyle 8n^3 + 12n^2 + 6n + 1 = (2n + 1)^3 is a composite number for all integers n.
    But to prove that it is a non-prime number we must show that it can be written as a product of two integers (x*y, where both x and y are different from 1)?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by mathnb View Post
    Thanks for the quick reply

    We can disprove the statement, i.e. prove that \displaystyle 8n^3 + 12n^2 + 6n + 1 = (2n + 1)^3 is a composite number for all integers n.
    But to prove that it is a non-prime number we must show that it can be written as a product of two integers (x*y, where both x and y are different from 1)?
    You are thinking along the right lines, but we won't need to go in that direction. If you like the x*y thing, you can stop at the second to last line, set each factor equal to 1, and solve for n. None will work.

    1 is not considered a prime number last I checked. And so, in light of this, no cube can be prime. How do we know this?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2010
    Posts
    4
    Got it. Thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: October 22nd 2011, 12:37 PM
  2. Replies: 1
    Last Post: June 19th 2011, 12:56 PM
  3. Replies: 3
    Last Post: February 17th 2011, 07:51 AM
  4. Replies: 6
    Last Post: August 27th 2010, 11:44 PM
  5. if ((2^n) -1 ) is prime, prove n is prime ?! >.<
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: March 18th 2009, 01:47 PM

Search Tags


/mathhelpforum @mathhelpforum