Prove that a Diophantine equation $ax^3 + by^2 = c$ has or doesn't have integer solutions

Question

I got a diophantine equation, specifically $5x^3 + 4y^2 = 535$ where I have to prove that there are or are not integer solutions.

I tried using modular artimetic, this is my process: $$4y^2 = 535 - 5x^3$$ $y^2$ mod 5 can be either 0, 1 or 4 (-1). $$4y^2 ≡ 0\textrm{ (mod 5)}$$ 5 divides the right hand side, so $y = 5t$. Furthermore: $$4(5t)^2 = 535 - 5x^3$$ $$100t^2 = 535 - 5x^3$$ $$20t^2 = 107 - x^3$$

But now I get stuck. 107 is a prime number and unfortunately its relatively large to deal with. I can rewrite this as $x^3 = 107 - 20t^2$, but I don't know where to proceed from here.

Answer 1

Let $X=-5x, Y=10y$ then we get Weierstrass form of the elliptic curve $Y^2 = X^3+13375$.
This curve has only two integral points, according to the magma online calculator as follows.

IntegralPoints($[0,0,0,0,13375]$);

It says that all integral points are $( -15 ,\pm 100)$.
See related info MO