• This solution requires basic knowledge of modular arithmetic.
• Take (mod $3$)

$x^2-12y+4=0\implies x^2=12y-4$

$12y-4\equiv 2(\text{mod} 3)$ 

$ \implies x^2\equiv 2(\text{mod} 3)$

Case 1: $x\equiv 0(\text{mod} 3)\implies x^2\equiv 0(\text{mod} 3)$

Case 2: $x\equiv 1(\text{mod} 3)\implies x^2\equiv 1(\text{mod} 3)$

Case 3: $x\equiv 2(\text{mod} 3)\implies x^2\equiv 1(\text{mod} 3)$

But, we know that $x^2\equiv 2(\text{mod} 3)$

So, this is a contradiction.

$\therefore$ There are no solutions to the given equation.