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Math Help - diophantine equation problem

  1. #1
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    diophantine equation problem

    Find all solutions to the equation x^2 + y^3 = z^2,\ x>0,\ y>0,\ z>0,\ (x,y,z)=1

    not a clue where to start with this. The book does not say anything about what to do when the exponents do not match and I can not seem to find anything on any of the various math help sites. Maybe I am just not knowing what to search for, but I am stuck. Can anyone point me in the right direction? In what way would this differ from having all exponents = 2?
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  2. #2
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    okay, so in one of those shower revelation moments, it occurred to me that the theory I was using earlier did not say that a square is a product of two squares...it said that a number to the nth power was a product of two numbers to the nth power. Using this glaringly obvious information I relooked at the problem and found:

     y^3 = z^2 - x^2 allows for either odd or even y. when z and x are both odd, y is even. When z and x are not both odd or even, then y is odd. When z and x are both even, y is even and (x,y,z)=2 and is a contradiction of (x,y,z)=1.

    So for this first case, where y is even, then z-x and z+x are both even. And if 2|y, then 2^3|y^3 and the math gives:
     y^3 = z^2 - x^2 = (z-x)(z+x), divide this through by 8 and you get
     (y/2)^3 = [(z-x)/4][(z+x/2)]=u^3v^3 setting (z-x)/4 = u^3 and (z+x)/2=v^3
    the math gives me
     <br />
x=v^3-2u^3,\ y=2uv,\ z=v^3+2u^3<br />

    The second case is when y is odd and z+x and z-x are both odd.
    So for this case the math gives:
     y^3 = z^2 - x^2 = (z-x)(z+x)=u^3v^3
    setting (z-x) = u^3 and (z+x)=v^3
    the math gives me
     x=(v^3-u^3)/2,\ y=uv,\ z=(v^3+u^3)/2

    The answer in the back of the book takes it a step further, which had me confused, by stating both answers in one combined format of:
     x=d|[(v^3-du^3)/2]|,\ y=duv,\ z=d[(v^3+du^3)/2] where either d=2, u>0, v>0, (u,v)=1, u odd or d=1, u>v>0, (u,v)=1, u odd, v odd.
    So my answers match up, I just need to take a look at the u odd vs u and v both odd when I am a little more awake! One more problem and I will only have one more lesson to go.
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  3. #3
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    y^3 = (z-x)(z+x)
    You divide by 8 and use fact that z-x,z+x are even.
    This is good. However, how do you know z-x is in fact divisible by 4? It can be z+x divisible by 4. It is one or the other.

    Thus we can have, (y/2)^3 = [(z-x)/4][(z+x)/2] or (y/2)^3 = [(z-x)/2][(z+x)/4].

    I do not think that really affects your answers so much but you need to consider that.
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