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Number theory proof?

  1. Sep 26, 2007 #1
    number theory problem

    For every prime [tex]p \ge 5[/tex], prove that [tex]p^2-1[/tex] is evenly divisible by 24(gives an integer answer).


    Example: for p=5, [tex]{5^2-1 \over 24} = 1[/tex]
     
    Last edited: Sep 26, 2007
  2. jcsd
  3. Sep 26, 2007 #2
    Notice that [tex]p^2 - 1 = (p - 1)(p + 1)[/tex]. If p is prime, try to examine how p - 1 and p +1 divide by 2, 3 and 4.
     
  4. Sep 26, 2007 #3
    woah, thats pretty neat, except for the 4 part you practically solved it it one setence. I'm finishing up my proof using modular arithmemtic.
     
  5. Sep 26, 2007 #4
    I see, your proof would say: for every prime [tex]p \ge 5[/tex], there is a k such that k is equal to to the integer remainder after the 2*2*2*3 have been cancelled out:

    [tex]k={p^2-1 \over 24} = {(p+1)(p-1) \over 24} = {(\text {even no.} \ge 6 = 2*3)(\text {even no.} \ge 4=2*2) \over 2*3*2*2}[/tex]

    so for example 7, would be

    [tex]k={7^2-1 \over 24} = {(7+1)(7-1) \over 24} = {(8)(6) \over 2*2*2*3} = {2*2*2*2*3 \over 2*2*2*3} = 2[/tex]

    My proof uses modular arithmetic, and mine simply says that for every prime [tex]p \le 5[/tex],

    [tex]p^2 \equiv 1 \text{ } (\text{mod }24) [/tex]

    so for example, for p=11

    [tex]11^2 = 121 [/tex]

    [tex]121 \equiv 1 \text{ } (\text{mod }24) [/tex]

    using modular arithmetic, we would read this as (121-1) is a multiple of 24, and indeed 120/24=5
     
  6. Sep 26, 2007 #5
    A theorem that might be useful is that every odd prime squared is congruent to 1 Mod 8. (In fact, every odd integer.)

    Proof: The integer is of the form 4n+r, where r is either 1 or 3. Then (4n+r)^2 = 16n^2+8n+r^2 = 8(M)+r^2. but r^2 = 1 or r^2 = 9.
     
    Last edited: Sep 26, 2007
  7. Sep 26, 2007 #6
    yeah, that's another proof. my proof can be directly deduced from yours.

    since you said for every odd prime squared,

    [tex]p^2 \equiv 1 \text{ } (\text{mod }8) [/tex]

    then mine directly follows as every odd prime squared, p>3, then

    [tex]p^2 \equiv 1 \text{ } (\text{mod }8*3) [/tex]
     
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