1. ## Diophantine equation

I have been trying to prove that $12\cdot{x^3}-3=n^2$ has no solutions in naturals numbers with $x>1$

I first thought of finding a mod in which it makes no sense, but I just couldn't find one (after some attempts).

2. Are we allowed to use Fermat's Last theorem?

$3\mid n$ and $n > 0$ since $x > 1$.

Let $n = 3m$ for some integer $m > 0$.

Then,
$12x^{3}-3 = 9m^{2}\Rightarrow 4x^{3}-1 = 3m^{2}$

Since the left-hand side is an odd number, $m$ is odd.

Let $m = 2k + 1$ for some integer $k\geq 0$.

By expanding, we get that
$4x^{3}= 3(2k+1)^{2}= 12k^{2}+12k+4$

Therefore, by dividing by four on both sides,
$x^{3}= 3k^{2}+3k+1\Rightarrow x^{3}+k^{3}= (k+1)^{3}$

If $k = 0$ , then $x = 1$ which is impossible since $x > 1$.

Otherwise, $k > 0$ and by Fermat's Last Theorem, there are no positive integer solutions to this equation.

P.S. I am sure that $x^{3}+k^{3}= (k+1)^{3}$ not having positive integer solutions can be proved using elementary means.

3. Originally Posted by fardeen_gen
P.S. I am sure that $x^{3}+k^{3}= (k+1)^{3}$ not having positive integer solutions can be proved using elementary means.
Fermat's Last Theorem: Fermat's Last Theorem: Proof for n=3