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Homework Help: Maximum common divisor

  1. May 21, 2006 #1
    Find all a(integers) that [tex](5a - 3:7a^2 - a + 1) = 1[/tex]

    I only know that

    [tex] d|7a^2 - a + 1 [/tex]
    [tex] d|5a - 3 [/tex]
     
  2. jcsd
  3. May 23, 2006 #2
    what is the meaning of "putting the 2 functions in the parathesis seperated by the colon" ??
     
  4. May 23, 2006 #3
    (a:b) = d
    d is maximum common divisor of a and b
     
  5. May 23, 2006 #4
    Correct me if I'm wrong, but aren't you asking for which values of a are the polynomials (5a - 3) and (7a^2 - a + 1) relatively prime? (5a - 3 : 7a^2 - a + 1) = 1 is what you have written. That doesn't quite mesh with your thread title so I'm a little lost. For example, where did "d" come from?
     
  6. May 23, 2006 #5

    Hurkyl

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    How do you normally find GCDs? The Euclidean algorithm, right? Have you tried it here?
     
  7. May 24, 2006 #6
    If (a:b) = 1 then a and b are co-prime
    Also (a:b) = c then c|a and c|b
    and exists d so that d|a , d|b and d|c
    thats the d I try to use.
     
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