1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Forehead slapper #1

  1. Apr 4, 2004 #1
    As a little diversion I thought i'd post this question which I call a forehead-slapper because that's what you'll likely do when you see the answer. You won't need more than high school maths to solve it.


    Show that if x and [tex]x^2+8[/tex] are primes then so is [tex]x^3-8[/tex]
     
  2. jcsd
  3. Apr 5, 2004 #2
    x = 3?

    cookiemonster
     
  4. Apr 5, 2004 #3

    Zurtex

    User Avatar
    Science Advisor
    Homework Helper

    Well [itex]x^3 - 8 = (x - 2)(x^2 + 2x + 4) [/itex] meaning that this can only be a prime when [itex]x = 3[/itex].

    Edit: I think I have proved the rest of it I'll let others have a go.
     
    Last edited: Apr 5, 2004
  5. Apr 5, 2004 #4
    There's a rest of it?

    cookiemonster
     
  6. Apr 5, 2004 #5

    Janitor

    User Avatar
    Science Advisor

    Given the Zurtex factorization, it seems pretty obvious that for x=4, 5, 6, ..., x^3-8 has to be composite, so I'm with Cookiemonster.
     
  7. Apr 5, 2004 #6
    Zurtex is exactly right. That was exactly the solution I had in mind. There is another one though I just realized, that doesn't involve having to factor a cubic.

    Note that any prime p except 3 is equal to 3k+1 or 3k-1 for some integer k. Then p^2 + 8 = 9k^2 ± 6k + 9, which is divisible by 3. 3^2+8=17, which is prime and also 3^3-8=19 is prime.
     
  8. Apr 5, 2004 #7

    Zurtex

    User Avatar
    Science Advisor
    Homework Helper

    Yes, the rest of it was obviously to prove that [itex]x[/itex] and [itex]x^2 + 8[/itex] could never both be prime. My proof was a little bit more complex that as it is early in the morning and I can't think simple maths yet :rolleyes:
     
  9. Apr 5, 2004 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The point of cookiemonster's post was that x= 3 is a prime number and that x2[/sup+8= 9+8= 17 is a prime number but x3= 27-8= 21= 3*7 is NOT.

    You can't prove your statement: it's not true.

    What Zurtex showed with "[itex]x^3 - 8 = (x - 2)(x^2 + 2x + 4) [/itex]" was that x3- 8 cannot be prime unless x= 3. That is essentially a converse of your original statement.
     
  10. Apr 5, 2004 #9

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    27-8 = 19 <filler space>
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Forehead slapper #1
  1. < or = to 1 (Replies: 5)

  2. -1 = 1? (Replies: 3)

  3. 1 = -1 ! (Replies: 16)

  4. -1 = 1 ? (Replies: 9)

  5. 1 = -1 ? (Replies: 7)

Loading...