Solving Diophantine equation without brute force I am having trouble trying to solve this Diophantine equations:
$$
^3=4^2+4−3
$$
I was wondering if anyone could help me find the integer solutions to these and any advice of techniques to use?
Thank you!
 A: The right hand side can be factored as
$$x^3=4y^2+4y-3=(2y+1)^2-4=(2y+3)(2y-1).$$
The gcd of the two factors on the right hand side divides their difference, which is $4$, but they are both odd so they are coprime. It follows from unique factorization that both factors are perfect cubes. But no two perfect cubes differ by $4$, a contradiction. Hence no integral solutions exist.
A: The equation in equivalent to 
$$
x^3+4=(2Y+1)^2=y^2,
$$
which is Mordell's equation. This has been solved here:
Solution to Mordell's Equation $y^2=x^3+4$
We see that $2Y+1=\pm 2$ is impossible, hence there is no solution.
A: Another way:  You can rewrite the equation: 
$$x^3+8=4y^2+4y+5$$
Or $$(x+2)(x^2-2x+4)=4y^2+4y+5$$
Let $(x+2;x^2-2x+4)=d (d\in N^+)$
Note that : $x^2-2x+4=x(x+2)-4(x+2)+12\Rightarrow d\mid 12$
We will prove $4y^2+4y+5\not\mid12$. Assume $4y^2+4y+5\mid12$, then we have $4y^2+4y+5=12k(k\in Z)$
$\Leftrightarrow (2y+1)^2=12k-4=4(3k-1)$
But $(2y+1)^2\equiv0;1\pmod{3}$ and $4(3k-1)\equiv2\pmod{3}$
Can you continue?
