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A fraction

  1. Feb 19, 2005 #1
    Hey, Guys...the following problem from IMO 1959, the very first: Prove that the fraction (21n+4)/(14n+3) is irreducible for every natural number n.
    ...just can't get what is meant by the given solution: 3(14n+3)-2(21n+4)=1...I mean, what is their "sample" method?!...can't see through...tried to get the same...no chance.
    Please, help!
     
  2. jcsd
  3. Feb 20, 2005 #2
    Showing that (21n + 4)/(14n + 3) is irreducible is equivalent to showing that 21n + 4 and 14n + 3 share no common factors, i.e. that gcd(21n + 4, 14n + 3) = 1 (where gcd = greatest common divisor). That, in turn, is equivalent to showing that there exists integers x, y such that x(21n + 4) + y(14n + 3) = 1 (which is what whoever wrote the solution did).

    One of the implications of the last theorem: if K divides both 21n + 4 and 14n + 3, then K divides x(21n + 4) + y(14n + 3), which implies that K divides 1. But then K = +/- 1, so gcd(21n + 4, 14n + 3) = 1.

    The proof of the other implication can be found http://www.math.swt.edu/~haz/prob_sets/notes/node7.html#SECTION00230000000000000000 [Broken].
     
    Last edited by a moderator: May 1, 2017
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