I want to find all the rational solutions of $x^2+y^3=z^3$

Question

I want to find all the rational solutions of $x^2+y^3=z^3$, and any help would be appreciated. My answer is as follows： $$x^2=z^3-y^3=(z-y)(z^2+zy+y^2)$$ Let $x=ab,a,b\in\Bbb Q$ when $$ z-y=a^2,z^2+zy+y^2=b^2$$ $$z^2+z(z-a^2)+(z-a^2)^2=b^2$$ Sort it out, and that's a Pell's equation: $$\left(\frac{2b}{a^2}\right)^2-3\left(\frac{2z}{a^2}-1\right)^2=1$$ Let $s=\frac{2b}{a^2},t=\frac{2z}{a^2}-1$ We obtained $$s^2-3t^2=1$$ The least $s_0=2,t_0=1$,other solutions can be obtained from: $$s_{n+1}=2s_n+3t_n,t_{n+1}=s_n+2t_n(n=0,1,2,...)$$ Last we obtained: $$x=ab=\frac{a^3}{2}s_n,z=\frac{a^2}{2}(t_n+1),y=\frac{a^2}{2}(t_n-1)$$

Example:Let $a=\frac{1}{3},s_1=7,t_1=4$,then $x=\frac{7}{54},y=\frac{1}{6},z=\frac{5}{18}$.

I don't know what the other solutions are,I'd be glad of some help with this,thank you for you help.

Answer 1

EDIT: Let's first consider the more symmetric equation $$x^2=y^3+z^3\tag1.$$If $y=0$, we see that gives the solutions $$(x,y,z)=(t^3,0,t^2)$$ for rational $t$. If $y\neq0$, we can set $x=ry$ and $z=ty$ with rational $r,t$. That gives $r^2=y(1+t^3)$, i.e. $r=0$, giving $t=-1$ and thus the solutions $$(x,y,z)=(0,y,-y),$$ or $y=r^2/(1+t^3)$, giving the solutions $$(x,y,z)=\left(\frac{r^3}{1+t^3},\frac{r^2}{1+t^3},\frac{r^2t}{1+t^3}\right).$$ That should take care of all possible cases.
To return from (1) to the original equation, we just have to change the sign of $y$, so the original equation has the solutions $$(x,y,z)=(t^3,0,t^2),$$ $$(x,y,z)=(0,y,y),$$ $$(x,y,z)=\left(\frac{r^3}{1+t^3},-\frac{r^2}{1+t^3},\frac{r^2t}{1+t^3}\right).$$

Answer 2

Well you can and draw another formula. $x^3+y^3=z^2$

$$x=(b^2-a^2)(b^2+2ba-2a^2)c^2$$

$$y=(b^2-a^2)(2b^2-2ab-a^2)c^2$$

$$z=3(b^2-a^2)^2(a^2-ab+b^2)c^3$$

equations $X^3+Y^3=Z^2$ Can be expressed by integers $p,s,a,b,c$ . where the number of $c$ characterizes the degree of primitiveness.

$$X=(3a^2+4ab+b^2)(3a^2+b^2)c^2$$

$$Y=2b(a+b)(3a^2+b^2)c^2$$

$$Z=3(a+b)^2(3a^2+b^2)^2c^3$$

And more.

$$X=2b(b-a)(3a^2+b^2)c^2$$

$$Y=2b(b+a)(3a^2+b^2)c^2$$

$$Z=4b^2(3a^2+b^2)^2c^3$$

If we decide to factor $X^3+Y^3=qZ^2$

For a compact notation we replace :

$$a=s(2p-s)$$

$$b=p^2-s^2$$

$$t=p^2-ps+s^2$$

then:

$$X=qb(a+b)c^2$$

$$Y=qa(a+b)c^2$$

$$Z=qt(a+b)^2c^3$$

And the most beautiful solution. If we use the solutions of Pell's equation: $p^2-3a^2s^2=1$

by the way $a$ May appear as a factor in the decision and. Then the solutions are of the form::

$$X=qa(2p-3as)sc^2$$

$$Y=q(p-2as)pc^2$$

$$Z=q(p^2-3aps+3a^2s^2)c^3$$

If we change the sign : $Y^3-X^3=qZ^2$

Then the solutions are of the form:

$$X=qa(2p+3as)sc^2$$

$$Y=q(p+2as)pc^2$$

$$Z=q(p^2+3aps+3a^2s^2)c^3$$

Another solution of the equation: $X^3+Y^3=qZ^2$

$p,s$ - integers asked us.

To facilitate the calculations we make the change. $a,b,c$

If the ratio is as follows : $q=3t^2+1$

$$b=2q(q+2\mp{6t})p^2+6p(t\mp1)ps+(q-1\mp{3t})s^2$$

$$c=6q(q-2(1\pm{t}))p^2+6q(t\mp1)ps+3(1\mp{t})s^2$$

$$a=12q(1\mp{t})p^2+6(4t\mp{q})ps+3(1\mp{t})s^2$$

If the ratio is as follows: $q=t^2+3$

$$b=3(q-1)(1\pm{t})s^2+2(3\pm{(q-1)t})ps+(1\pm{t})p^2$$

$$c=3(6-(q-1)(q-3\mp{t}))s^2+6(1\pm{t})ps+(q-3\pm{t})p^2$$

$$a=3(6-(q-1)(1\mp{t}))s^2+6(1\pm{t})ps+(1\pm{t})p^2$$

Then the solutions are of the form:

$$X=2c(c-b)$$

$$Y=(c-3b)(c-b)$$

$$Z=3a(c-b)^2$$

Then the solutions are of the form:

$$X=2(c-b)c$$

$$Y=2(c+b)c$$

$$Z=4ac^2$$

If the ratio is as follows : $q=t^2+3$

$$c=6(q-4)(2\pm{t})p^2+4(6\pm{(q-4)t})ps+2(2\pm{t})s^2$$

$$b=3(24-(q-4)(q-3\mp{2t}))p^2+12(2\pm{t})ps+(q-3\pm{2t})s^2$$

$$a=3(24-(q-4)(4\mp{2t}))p^2+12(2\pm{t})ps+2(2\pm{t})s^2$$

If the ratio is as follows $q=3t^2+4$

$$c=q(-q+7(4\mp{3t}))p^2+6q(t\mp{1})ps+(q-4\mp{3t})s^2$$

$$b=3q(2q-7(1\pm{t}))p^2+6q(t\mp{1})ps+3(1\mp{t})s^2$$

$$a=21q(1\mp{t})p^2+6(7t\mp{q})ps+3(1\mp{t})s^2$$

Then the solutions are of the form :

$$X=2(3c-2b)c$$

$$Y=2(3c+2b)c$$

$$Z=12ac^2$$

Then the solutions are of the form :

$$X=(2b-c)b$$

$$Y=(2b+c)b$$

$$Z=2ab^2$$