How to prove $~\sqrt{3}\sqrt{4a^{3}-1}~$ isn't an integer? I'm trying solve: $~a^3 + b^3 = c^3~$ has no nonzero integer solutions.
If $~(c−b)=1~$ then $~c^3-b^3=3c^2-3c+1=a^3,~$
from  Wolframalpha get:
$$
c = \dfrac{3- \sqrt{3}\sqrt{4a^{3}-1}}{6} \\
c = \dfrac{\sqrt{3}\sqrt{4a^{3}-1}+3}{6}
$$
How to prove $~\sqrt{3}\sqrt{4a^{3}-1}~$ isn't an integer? (eidt: while $~a,\ b,\ c ~$ are nonzero integers)
 A: You have to prove that
$$ 12 a^3 = b^2+3 \tag{1}$$
has no integer solutions. Since $3\mid$LHS, we must have $b=3c$, hence:
$$ 4 a^3 = 3c^2+1,\tag{2}$$
and $c$ must be odd, hence $c=2d+1$ and:
$$ a^3 = 3 d^2 + 3 d + 1 = (d+1)^3-d^3.\tag{3}$$
So you just have to prove that the difference between two consecutive cubes is never a cube (except for $1^3-0^3=1^3$). This is usually achieved by splitting the RHS of $(3)$ over $\mathbb{Z}[\sqrt{-3}]$ and by using the fact that the ring of Eisenstein integers $\mathbb{Z}[\omega]$ is an euclidean domain, hence a unique factorization domain. We can use Fermat's descent too, since $(3)$ implies $a=3e+1$ and:
$$ 3e(3e^2+3e+1)= d(d+1).\tag{4}$$
A: Multiplying with $\sqrt{3}$ only returns an integer when multiplying with certain multiples of three.
Assuming $~4a^3 \equiv 1 (3) \equiv 4(3)$ we check for divisibility by three:
$$4a^3 \equiv4(3) \iff \exists k \in \mathbb N:4a^3-4=3k$$
$$3|4(a^3-1) \iff 3|a^3-1 \iff \Big( 3|a^{3/2}-1 ~~~~\vee~~~~ 3|a^{3/2}+1 \Big)$$
That is not true for $\forall a\not = 1$ and thus your product cannot be an integer.
