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n**n + (n+1)**(n+1) = 0 mod p**2

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  • Kermit Rose
    Suppose n and p are such that n**n + (n+1)**(n+1) = 0 mod p**2 then n**n = - (n+1)**(n+1) mod p**2 Since for n 0, n and n+1 are always relatively prime, p
    Message 1 of 1 , Sep 9 12:46 PM
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      Suppose n and p are such that

      n**n + (n+1)**(n+1) = 0 mod p**2

      then

      n**n = - (n+1)**(n+1) mod p**2

      Since for n > 0, n and n+1 are always relatively prime,

      p cannot divide both n and (n+1).

      Thus from the equation

      n**n = - (n+1)**(n+1) mod p**2


      p cannot divide either of n or (n+1).

      Divide by (n+1)**n

      (n/(n+1)) ** n = -(n+1) mod p**2


      How will the order of (n/(n+1)) mod p**2 and the order of (-(n+1)) mod
      p**2 relate to each other?

      Kermit
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