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Finding unions of sets

  1. Mar 5, 2009 #1
    1. The problem statement, all variables and given/known data
    X = {x: x=4n+1, n[tex]\in[/tex][tex]N[/tex]}
    Y = {y: y=m[tex]^{2}[/tex]+m+1, m[tex]\in[/tex][tex]N[/tex]}

    Find: Y[tex]\cap[/tex]X

    3. The attempt at a solution

    First I tried to set x=y
    [tex]\Rightarrow[/tex] n[tex]^{2}[/tex]+n+1=4n+1
    [tex]\Rightarrow[/tex] n[tex]^{2}[/tex]-3n=0
    [tex]\Rightarrow[/tex] n(n-3)=0
    [tex]\Rightarrow[/tex] n=0,3

    Subbing this in to either equation yields {1,13}.
    I thought I was doing okay.
    However I was comparing assignments with a classmate and she had this massive set that just keeps on going: {1, 13, 21, 57, 73, 133, 157, 241, 273, ...}
    After I thought about it, this made sense, as my solution only found the elements that are in both sets when n is the same for both equations, but set equality relies only on its elements, not at what stage they were reached.
    She'd done this the long way round, scripting a little program to calculate all elements for both sets then compare the elements and find the common ones.
    What I'd like to know is, how can I go about solving this algebraically?

    Thanks in advance,
    Yorick.
     
  2. jcsd
  3. Mar 6, 2009 #2
    write the equation as it is.

    m^2 + m + 1 = 4n + 1
    --> m^2 + m = 4n
    --> m(m+1) = 4n

    now rhs is a multiple of 4... and therefore, so must the lhs. But if m is even, m + 1 is odd...
    do you see where i am going with this? can you proceed?
     
  4. Mar 9, 2009 #3
    Yes, Thankyou!

    So basically the solution is something like {x: x=m^2 + m + 1 ; m/4[tex]\in[/tex][tex]N[/tex] or (m+1)/4[tex]\in[/tex][tex]N[/tex]}
     
  5. Mar 10, 2009 #4
    yep... thats right!
     
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