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Hard task with primes

  1. Sep 26, 2007 #1
    Find all triples of primes (p,q,r), that pq+qr+rp and p^3+q^3+r^3−2pqr are divisible by p+q+r.
    I really don't know how to start, (of course i've been trying)
  2. jcsd
  3. Sep 26, 2007 #2


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    Hint: start by multiplying out (p+q+r)(p2 + q2 + r2)
  4. Sep 26, 2007 #3


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    By "p2+ q2+ r2" do you mean p^2+ q^2+ r^2 ?
  5. Sep 26, 2007 #4
    it doesn't simplify those quotients( I was counting about an hour)
  6. Sep 26, 2007 #5
    please, help me:)
  7. Sep 26, 2007 #6


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    Sorry about the typo!

    [itex](p+q+r)(p^2 + q^2 + r^2)[/itex]
    [itex]= p^3 + q^3 + r^3 + pq(p+q) + qr(q+r) + rp(r+p)[/itex]
    [itex]= p^3 + q^3 + r^3 + (p+q+r)(pq+qr+rp) - 3pqr[/itex]
    [itex]= (p^3 + q^3 + r^3 - 2pqr) + (p+q+r)(pq+qr+rp) - pqr[/itex]

    The whole expression is divisible by (p+q+r)
    If the given conditions hold, pqr is divisible by p+q+r.

    But p,q,r are primes, therefore...

    Notes: the question doesn't say p,q,r are distinct.
    And so far, we haven't used the fact that p+q+r divides pq +qr + rp.
  8. Sep 27, 2007 #7
    ok, now i see your solution, thanks, big thanks
  9. Nov 4, 2007 #8
    Hey! It's task from Polish Olympiad in Mathematics 2007/2008. Please, delete this thread. And shame on you, menager31!
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