Page 1 of 3 123 LastLast
Results 1 to 10 of 22

Thread: find all solutions to 2p + 1 = n^2, for n natural, p prime

  1. #1

    find all solutions to 2p + 1 = n^2, for n natural, p prime

    if n is a natural number and p is a prime number find all solutions to 2p + 1 = n^2

  2. #2
    Elite Member mmm4444bot's Avatar
    Join Date
    Oct 2005
    Location
    Seattle
    Posts
    7,583

    Re: solutions to 2p + 1 = n^2



    Did I already give you my boilerplate response for posters who don't show any work, don't explain what they already know, and don't ask any questions?

    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  3. #3

    Re: solutions to 2p + 1 = n^2

    no, but I figured I would get an answer like that, so you really don't have to.

    My question is.... where do I begin? My professor said he had not taught us what we needed to know to fully complete the problem (and therefore made it extra credit)

    would the division algorithm be used? (a= qb + r)

  4. #4
    Elite Member mmm4444bot's Avatar
    Join Date
    Oct 2005
    Location
    Seattle
    Posts
    7,583

    Re: solutions to 2p + 1 = n^2

    Quote Originally Posted by twisted_logic89

    My question is.... where do I begin?


    2p = n[sup:391bq5es]2[/sup:391bq5es] – 1

    2p = (n + 1) * (n – 1)

    If n is a natural number, what can we say about the factors of 2p above?

    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  5. #5

    Re: solutions to 2p + 1 = n^2

    in n is a natural number then the factors of 2p would have to be greater than zero, right?

  6. #6
    Senior Member
    Join Date
    Jan 2006
    Posts
    1,307

    Re: solutions to 2p + 1 = n^2

    Quote Originally Posted by twisted_logic89
    in n is a natural number then the factors of 2p would have to be greater than zero, right?
    You knew that from the assumption on p. In fact, I'd venture to say 2p >= 4

  7. #7
    Elite Member
    Join Date
    Jun 2007
    Posts
    12,635

    Re: find all solutions to 2p + 1 = n^2, for n natural, p prime

    [tex]p = \frac{1}{2}\cdot (n+1)\cdot (n-1)[/tex]

    Now my question is -

    What is the definition of a prime number?

    Then my request is - after you found the definition - sit and think about the step above.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  8. #8
    Elite Member mmm4444bot's Avatar
    Join Date
    Oct 2005
    Location
    Seattle
    Posts
    7,583

    Re: solutions to 2p + 1 = n^2

    Quote Originally Posted by twisted_logic89

    [If] n is a natural number then the factors of 2p would have to be greater than zero, right?


    If you think of the expression 2p as twice some prime number, then my answer to your question is clearly "yes".

    If you think of the expression 2p as n[sup:1278fexe]2[/sup:1278fexe] - 1, for any natural number n, then my answer to your question is "not always".


    [tex]\begin{tabular}{|c|c|c|c|} \hline n & n - 1 & n + 1 & 2p & \hline 1 & 0 & 2 & 0 & 2 & 1 & 3 & 3 & 3 & 2 & 4 & 8 & 4 & 3 & 5 & 15 & 5 & 4 & 6 & 24 & : & : & : & : & \hline \end{tabular}[/tex]


    Subhotosh gave you the important question to consider; here are some more questions to consider.

    How many prime numbers are even?

    Is the value of 2p even or odd when p is any real number?

    Is the product (n - 1) * (n + 1) even or odd when n is a natural number?

    When n is odd, can you factor out a number from the product (n - 1) * (n + 1)?

    If so, then can you simplify the original equation when n is odd?

    ~ Mark

    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  9. #9

    Re: find all solutions to 2p + 1 = n^2, for n natural, p prime

    prime number- divided evenly only by itself and 1

    the only even prime number is 2

    2p will be even

    product of (n-1)*(n+1) is even

    so... one could factor out a 2 from the original equation, correct?

    p= (1/2) * (n+1) * (n-1)

    ! i feel like I am getting really close with this...
    are the solutions for this 1 and 2?

  10. #10
    Elite Member mmm4444bot's Avatar
    Join Date
    Oct 2005
    Location
    Seattle
    Posts
    7,583

    Re: find all solutions to 2p + 1 = n^2, for n natural, p prime

    Quote Originally Posted by twisted_logic89

    ... are the solutions for this 1 and 2?


    Hmmm, it seems to me that stating the numbers 1 and 2 as solutions for an equation that contains two different symbols is somewhat ambiguous.

    If you intend these two numbers to be p and n, then I suppose that you mean n = 1 and p = 2.

    n = 1

    n[sup:2rg8x8lu]2[/sup:2rg8x8lu] = 1

    Is there a prime number p such that 2p + 1 = 1?

    I don't think so. (It's certainly not p = 2.)

    Maybe you meant n = 1 and n = 2 as solutions.

    n = 2

    n[sup:2rg8x8lu]2[/sup:2rg8x8lu] = 4

    Is there a prime number p such that 2p + 1 = 4?

    I don't think so.

    I don't think 1 and 2 are solutions.

    Keep trying.

    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •