1. Not finding help here? Sign up for a free 30min 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!

Which primes p satisfy p^2|5^p^2+1?

  1. Nov 13, 2009 #1
    1. The problem statement, all variables and given/known data

    First of all, hi everyone!
    I'm quite new in number theory, and need help on this one badly...

    Determine all prime numbers p so p2 divides 5p2+1.
    2. Relevant equations

    Euler's theorem: If a and m are coprimes then [tex]a^{\varphi(m)}\equiv 1 (mod\ m)[/tex]
    where [tex]\varphi(m)[/tex] (Euler's function) denotes number of positive integers which are coprime with m and not greater than given int m.

    Special: Fermat's little theorem... if p is prime, p and a coprimes, then [tex]a^{p-1}\equiv 1 (mod\ p)[/tex]

    ...and... [tex]\varphi(p^{2})=p^{2}-p[/tex]

    3. The attempt at a solution

    Know one solution p=3, but I got it by assumption. :((

    Thanks in advance!
     
  2. jcsd
  3. Nov 14, 2009 #2

    Hi there (=
    I'm also new to number theory.
    May i check with you if the primes that satisfy the above condition is 2 and 3?
    I am writing my full proof as i am living in(asia) different GMT from yours.






    <Question : Determine all prime numbers p so p2 divides 5p2+1.>

    By Fermat's Little Theorem,

    [tex]5^p[/tex] is congruent to 5 (mod p) (1)

    Which suggest that [tex]5^{p.p} = 5^{p}^{2}[/tex] is congruent to 5 (mod p) as stated in (1).

    Since [tex]5^{p}^{2}+1 [/tex]is divisible to [tex] p^{2}[/tex] ;

    Therefore, [tex]5^{p}^{2}+1 [/tex] is divisible to p ;

    5+1 (mod p) is congruent to 0 (mod p)

    Thus, 6 is a multiple of p .

    pk = 6=2*3 with k is an element of integer.

    By Euclid's Lemma,

    Therefore , p can be either 2 or 3.
     
    Last edited: Nov 14, 2009
  4. Nov 14, 2009 #3
    Well, thanks a lot, it seems correct!
    Meanwhile, I did it too...
    According to Euler's Theorem:
    [tex]5^{p^{2}-p}\equiv 1(mod\ p^{2})[/tex]
    [tex]5^{p^{2}}\equiv 5^{p}(mod\ p^{2})[/tex]
    [tex]5^{p^{2}}-5^{p}\equiv 0(mod\ p^{2})[/tex]
    [tex]5^{p^{2}}+1-(5^{p}+1)\equiv 0(mod\ p^{2})[/tex]
    So, if [tex]5^{p^{2}}+1[/tex] is divisible by p2, then [tex]5^{p}+1[/tex] must be, too.
    Now, using Little Fermat's Theorem:
    [tex]5^{p-1}\equiv 1(mod\ p)[/tex]
    [tex]5^{p}\equiv 5(mod\ p)[/tex]
    [tex]5^{p}-5\equiv 0(mod\ p)[/tex]
    [tex]5^{p}+1-6\equiv 0(mod\ p)[/tex]
    If [tex]5^{p}+1[/tex] is divisible by p, then 6 must be divisible by p. The only prime numbers which satisfy this are, as you proved, p=2 and p=3. But, if we check it with 1st statement, we'll see that p=3 is unique solution in N field.

    Thank you once again, cheers! :))
     
  5. Nov 15, 2009 #4

    Mentallic

    User Avatar
    Homework Helper

    For p=2,
    [tex] p^2|5^p^2+1[/tex]
    is 156.5

    Now, I haven't studied number theory, but isn't the definition of "divides" that the solution be a whole number? Which, as you can see for p=2 it is not.
     
  6. Nov 16, 2009 #5
    yes you are right i've forgotten to check with the first statement (=
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook