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)

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.
