Diophantine Equation solved with elliptic curves

Question

I want to know how to find all solutions in $\mathbb{Z}$ for $$ 2a^2 -3ab +5c^2 =0. $$ I already solved it and I will post my solution soon. One solution for example is $(15,11,3).$

Answer 1

We substitute $x=\frac{a}{b}$ and $y=\frac{c}{b}$ and find all the intersections of $2 x^{2}-3 x+5 y^{2}=0$ and $y=t x$. We find $$ 0=2 x^{2}-3 x+5 t^2 x^{2}=x\left(\left(2+5 t^2\right) x-3\right) . $$ So the intersection points are $(0,0)$ and $$ \left(\frac{3}{2+5t^2}, \frac{3t}{2+5t^2}\right). $$ The second intersection is in $\mathbb{Q}^{2}$ exactly when $t \in \mathbb{Q}$. Now we write $t=\frac{n}{m}$ with $m, n \in \mathbb{Z}$ and $\operatorname{gcd}(m, n)=1$, and get $$ \left(\frac{3 m^{2}}{2 m^{2}+5 n^{2}}, \frac{3mn}{2 m^{2}+5 n^{2}}\right). $$ Since $x=\frac{a}{b}$ and $y=\frac{c}{b},$ we see that each solution $(a, b, c)$ is a (rational) multiple of $$ \left(3 m^{2}, 2 m^{2}+5 n^{2}, 3 m n\right), $$ where $m, n \in \mathbb{Z}$ and $\operatorname{gcd}(m, n)=1$.

What are the primitive solutions? We have $\operatorname{gcd}\left(3 m^{2}, 3 m n\right)=3 m$. Can 3 be a divisor of $2 m^{2}+5 n^{2}$? We have $$2 m^{2}+5 n^{2}=2\left(m^{2}+n^{2}\right) \bmod 3$$

and $$ \begin{aligned} m^{2} &\equiv 0 \bmod 3\\ n^{2} &\equiv 1 \bmod 3. \end{aligned} $$ So 3 divides $2 m^{2}+5 n^{2}$, exactly when $3 \mid m, n$, but $\operatorname{gcd}(m, n)=1$, Contradiction.

Now $\operatorname{gcd}\left(3 m, 2 m^{2}+5 n^{2}\right)=\operatorname{gcd}\left(m, 5 n^{2}\right)=\operatorname{gcd}(m, 5)$ holds. So the primitive solutions with $b \neq 0$ are : $$ (a, b, c)=\frac{1}{d}\left(3 m^{2}, 2 m^{2}+5 n^{2}, 3 m n\right) \text { for } \operatorname{gcd}(m, n)=1 \text { and } d=\operatorname{gcd}(5, m) \text {. } $$ Note: For $(m, n)=(0,1)$, $d=5$ and $(a, b, c)=(0,1,0)$. For $b=0$ there is only the solution $(a, b, c)=(0,0,0)$, i.e. $(m, n)=(0,0)$.