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Homework Help: 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


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    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 (=
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