Solving the diophantine $y^3=2x^2$

Question

I am trying to find the general solutions to the diophantine $y^3=2x^2$. Usually I can solve these using coprime/prime factorisation methods but I have tried and cannot seem to get the correct general solutions.

My efforts so far:

$2x^2=2^{r_0+1} \times p_1^{2r_1}\times p_2^{2r_2}......p_k^{2r_k}=2^{3s_0}\times q_1^{3s_1}\times q_2^{3s_2}\times ........q_l^{3s_l}=y^3$

Then I wanted to use the fundamental theorem of arithmetic, so $2r_0+1=3s_0$ and for all $i\gt0$, $2r_i=3s_i$.

Although now I do not know how to carry on with finding the general solutions? Can anyone point me in the right direction? Thank you!

Edit: Trying to get into the form: $(2^3,2^2)$ for $n \in \mathbb(Z)$ (FOR EXAMPLE)

Answer 1

Clearly $x$ and $y$ are even, thus we can replace them by $x=2a, \ y= 2b$ . After dividing both sides by 8 we get the equation $b^3=a^2$. Now can you solve it using unique factorization?

EDIT: $2 | 2x^2$, thus $2 | y^3$, we can write $y=2b$, substituting into our equation we get $8b^3 = 2x^2$ thus $4b^3= x^2$ hence $x$ is even, write $x=2a$, substitute again, and you should get $b^3 = a^2$, now by unique facotrization $a$ is a cube and $b$ is a square, thus $(a, b) = (n^3, n^2)$, since we're solving for $x$ and $y$, $ \ \ (x, y) = (2a, 2b) = (2n^3, 2n^2)$, where $n \in \mathbb{Z}$

Answer 2

Collecting up the comments:

For any $n$ let $$ x = 2n^3, \quad y = 2n^2 . $$ Clearly $$ y^3 = 8n^6 = 2x^2 . $$ You can prove these are all the solutions using the prime factorizations. Any odd prime $p$ that divides both $x$ and $y$ must appear $3r$ times in $x$ and $2r$ times in $y$ so use $p$ as a factor of $n$.

Answer 3

$y=2y_1\Rightarrow8y_1^3=2x^2\iff y_1^3=\left(\dfrac x2\right)^2$. Put now $y_1=t^2$ and $x=2t^3$ so you have the identity $t^6=t^6$ from which the parameterization of $x$ and $y$