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Another equation

  1. Jan 30, 2007 #1
    how to find the integer solutions of the following equation:
    [tex]x^3 + x^{2}y + xy^2 + y^3 = 8(x^2 + xy + y^2 + 1)[/tex]

    after simplifying, i got,
    [tex](x^2 + y^2)(x + y - 8) = 8(xy + 1)[/tex]

    from this, i can see that x and y are both odd or both even. but what should i do next. i tried putting x = 2p and y = 2q but that's not going anywhere.

    thanks in advance.
     
  2. jcsd
  3. Jan 31, 2007 #2
    anyone? please:confused:
     
  4. Feb 1, 2007 #3
    Murshid ,my man,here it is

    Obviously [itex]x\equiv y (mod2)[/itex].
    WLOG,put x=u+v,y=u-v and it is sufficient to consider:
    [tex](u^2+v^2)(u-4)-2(u^2-v^2+1)=0[/tex]
    • if v=0,then u3-6u2-2=0
      This equation has no solutions in [itex]\mathbb{Z}[/itex]
    • if u=0,then v2+1=0 has no solutions in [itex]\mathbb{Z}[/itex]
    • if [itex]|u|\geq 1,|v|\geq 1[/itex] ,then, [itex] u^2+v^2>|u^2-v^2+1|[/itex] with consequence that [itex]|u-4|=1[/itex] must hold.
    • if u=3 ,than v2-29=0 has no solutions in [itex]\mathbb{Z}[/itex]
    • if u=5 ,than v2-9=0 has solutions v1=3,v2=-3

    Thus, the diophantine equation we started with, has only two solutions in x,y :
    S={(2,8),(8,2)}
     
    Last edited: Feb 1, 2007
  5. Feb 2, 2007 #4
    i didn't understand a few things. for example,

    where is this used?

    how did you get this?
     
  6. Feb 3, 2007 #5
    ok, i understood this much:
    if x = u+v and y = u-v, then x - y = 2v. therefore,
    [tex]x - y \equiv 0 (\bmod \\ 2)[/tex]
    [tex]x \equiv y (\bmod \\ 2)[/tex]

    but what about the second one i mentioned in my last post? what i want to know is how did tehno get:

     
    Last edited: Feb 3, 2007
  7. Feb 3, 2007 #6
    answer

    by splitting it in two cases:

    a) [itex]|u|\geq|v|\geq 1\rightarrow u^2-v^2+1\leq u^2<u^2+v^2[/itex]

    b) [itex]|v|>|u|\geq 1\rightarrow |u^2-v^2+1|=v^2-u^2-1<v^2<u^2+v^2[/itex]

    Joining these two in other words means that:

    [tex]|u-4|=|2(u^2-v^2+1):(u^2+v^2)|<2\longrightarrow|u-4|=0\ or \ |u-4|=1[/tex]
    which I specified in the post .
     
    Last edited: Feb 3, 2007
  8. Feb 4, 2007 #7
    is this how you got a) and b) ?:
    [tex]|u| \geq |v| \geq 1[/tex]

    [tex]u^2 \geq v^2 \geq 1[/tex]

    [tex]-u^2 \leq -v^2 \leq -1[/tex]

    [tex]0 \leq u^2 - v^2 \leq u^2 -1[/tex]

    [tex]1 \leq u^2 - v^2 + 1 \leq u^2[/tex]

    now [tex]u^2 - v^2 + 1 \leq u^2 < u^2 + v^2[/tex]

    again,
    [tex]|v| > |u| \geq 1[/tex]

    [tex]v^2 > u^2 \geq 1[/tex]

    [tex]-v^2 < -u^2 \leq -1[/tex]

    [tex]0 < v^2 - u^2 \leq v^2 - 1[/tex]

    [tex]1 < v^2 - u^2 + 1 \leq v^2[/tex]

    now,
    [tex]v^2 - u^2 - 1 < v^2 - u^2 + 1 \leq v^2 < u^2+ v^2[/tex]
    [tex]v^2 - u^2 - 1 < v^2 < u^2 + v^2[/tex]

    but how did you get the following:
     
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