show that $a^3+4a=b^2$ then exist $M\in\mathbb{N}$ s.t $a=2M^2$ I would appreciate if somebody could help me with the following problem: 
Q: show that 
If $a,b$ satisfy equation 
$$a^3+4a=b^2(a,b\in\mathbb{N})$$ then there exist $M\in\mathbb{N}$ s.t $a=2M^2$

Let $d = \gcd(a,b)$. So $a = xd$, and $b = yd$. Thus: $x^3d^3 + 4xd = y^2d^2$. So $d^2x^3 + 4x = dy^2$. This gives: $x$ divides $dy^2$. But $\gcd(x,y) = 1$ so $\gcd(x,y^2) = 1$. Thus $x$ divides $d$. So $d = kx$. So: $k^2x^5 + 4x = kxy^2$. Thus: $k^2x^4 + 4 = ky^2$. Hence: $4 = k(y^2 - kx^4)$. This means $k$ divides $4$. So $k = 1, 2, 4$. 
Case 1: $k = 1$. So: $4 = y^2 - x^4 = (y - x^2)(y + x^2)$. So $y - x^2 = 1, y + x^2 = 4$. So $2y = 5$. This can't happen since $y$ has to be an integer. 
Case 2: $k = 2$. So: $2 = y^2 - 2x^4$. So $y^2 = 2(x^4 + 1)$. So $x^4 + 1 = 2m^2$. Observe that this equation has initial solution: $x = 1 = m$ and it will have infinitely many solutions.
Case 3: $k = 4$. So: $1 = y^2 - 4x^4 = (y - 2x^2)(y + 2x^2)$. This can't happen. So we have: $k = 2$ is the only acceptable solution and this means: $a = 2x^2 = 2M^2$ with $M = x$. Done.
