Diophantine Equation solved with elliptic curves 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 $(15,11,3).$

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. $(m, n)=(0,0)$.
