The question I was given: Find all solutions to $x^2 + y^3 = z^2$ in which x,y, and z are pairwise relatively prime and y is even. What I have so far: If $y$ is even then the equation becomes $x^2+8*b^3=z^2$ where b is the integer $y/2$ and $x$ and $z$ must be odd. I don't know if I should do anything mod 8 or how to approach this problem.
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4I'd write $y^3=z^2-x^2$ since both sides factor. – Angina Seng May 19 '18 at 15:15
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Note that every number not $\equiv 2 \pmod 4$ can be written as the difference of $2$ squares. If $y$ is odd, for example, then you could take $(x,y,z)=\left(\frac {y^3+1}2,y,\frac {y^3-1}2\right)$. – lulu May 19 '18 at 15:22
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This mentions the equation Solvability of the Diophantine equation $x^{2} - y^{2} = 4z^{n}$?. Also here an answer gives some parametric solutions Find all pairs of positive integers $(x,y)$ such that $x^2-y^2=a^3$ and $x^3-y^3=b^2$ for integral $a, b$? – Sil May 19 '18 at 15:53
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Above equation shown below;
$x^2 + y^3 = z^2$
Since variable $(y)$ is required to be even & comment
given by "Lulu" gives rational solution of $(x,z)$ instead of integer solution,
another solution is given below:
$(x,y,z)= ((2k^2-k),(2k),(2k^2+k))$
For $k=2$ we get $(x,y,z)=(6,4,10)$
Sam
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