Number theory

Home Forums Math Olympiad Number theory

Viewing 3 posts - 1 through 3 (of 3 total)
  • Author
    Posts
  • #69045
    divyanshu gupta
    Participant

    Please help in 13,19 25 27 29

    Attachments:
    You must be logged in to view attached files.
    #69880
    divyanshu gupta
    Participant

    Please help

    #70270
    Shirsendu Roy
    Moderator

    13

    p congruent to 3(mod 4)

    given 4|(p-3)

    or, p-3=4k

    or, p=4k+3 k is in set of integers

    or, p-1=4k+2=2(2k+1) which is given equation

    by fermat’s theorem

    2^{p-1}4^{p-1}6^{p-1}….(p-1)^{p-1}is congruent to 1(mod p)

    or, {(2)(4)(6)….(p-1)}^{p-1}is congruent to 1(modp)

    or,{(2)(4)(6)…..(p-1)} is congruent to 1^{\frac{1}{p-1}}(modp)

    or, {(2)(4)(6)….(p-1)} is congruent to 1^{\frac{1}{2(2k+1)}(mod p)

    or, {(2)(4)(6)….(p-1)} is congruent to =+-1(modp)

Viewing 3 posts - 1 through 3 (of 3 total)
  • You must be logged in to reply to this topic.