# Thread: diophantine equation problem

1. ## diophantine equation problem

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

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?

2. 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:

$\displaystyle y^3 = z^2 - x^2$ 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:
$\displaystyle y^3 = z^2 - x^2 = (z-x)(z+x)$, divide this through by 8 and you get
$\displaystyle (y/2)^3 = [(z-x)/4][(z+x/2)]=u^3v^3$ setting $\displaystyle (z-x)/4 = u^3$ and $\displaystyle (z+x)/2=v^3$
the math gives me
$\displaystyle x=v^3-2u^3,\ y=2uv,\ z=v^3+2u^3$

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

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:
$\displaystyle x=d|[(v^3-du^3)/2]|,\ y=duv,\ z=d[(v^3+du^3)/2]$ 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.

3. $\displaystyle y^3 = (z-x)(z+x)$
You divide by $\displaystyle 8$ and use fact that $\displaystyle z-x,z+x$ are even.
This is good. However, how do you know $\displaystyle z-x$ is in fact divisible by $\displaystyle 4$? It can be $\displaystyle z+x$ divisible by $\displaystyle 4$. It is one or the other.

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

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