<s>[INST] Proving there are no integer solutions for $3x^2=9+y^3$ 
Prove there are no $x,y\in\mathbb{Z}$ such that $3x^2=9+y^3$.

Initial proof
Let us assume there are $x,y\in\mathbb{Z}$ that satisfy the equation, which can be rewritten as $$3(x^2-3)=y^3.$$ So, $$3 \mid y \Rightarrow 3^3 \mid y^3 \Rightarrow 3^2 \mid x^2 - 3.$$ As $3 \mid -3$, it follows that $$3 \mid x^2 \Rightarrow 3^2 \mid x^2.$$ Say $x^2=3^2 \cdot a^2$ for some $a \in \mathbb{Z}$. Then $$\begin{align*}x^2-3 &= 3^2 \cdot a^2 - 3 \\&= 3 \cdot  (3 \cdot a^2 - 1)\end{align*}$$ As $3^2 \mid x^2-3$, it follows that $3 \mid 3 \cdot a^2 - 1$. It is obvious that $3 \mid 3 \cdot a^2$, so it follows that $3 \mid -1$, which is false. Therefore, the assumption that there are $x,y\in\mathbb{Z}$ was wrong.
Alternative proof
$$
\begin{align*}
3x^2&=9+y^3\\
3(x^2-3)&=y^3
\end{align*}
$$
So, $3 \mid y \Rightarrow 3^3 \mid y^3 \Rightarrow 3^2 \mid x^2 - 3 \Rightarrow 3^2 \mid x^2 \Rightarrow 3^2 \not\mid x^2 - 3$ and we are done.

Both proofs are essentially the same, except for the fact that the second proof is much shorter. I think the first proof is much more clear, but it takes a bit longer to write down.


*

*Would the second proof suffice in a math competition, or is it too short?

*This problem was part of the Dutch finals of 1978. I figured it out in under a minute, which is quite unusual for me, so now I wonder: is my proof correct?

 [/INST] Assume that the expression has integer solutions, or atleast one for that matter: 
Work in modulus $3$, if: 
$$3(x^2-3)=y^3$$ 
$y^3$ must be a multiple of $3$. Then we can set $y=3k$ where $k$ is an integer. Now we have: 
$$27k^3=3(x^2-3)$$
Dividing both sides by $3$ gives us $9k^3=x^2-3$
Taking $-3$ to the other side shows us that $x^2$ is a mutiple of $3$. Thus $x$ is a multiple of $3$. Then we can set $x=3t$ where $t$ is an integer. This gives us: 
$$9k^3+3=9t^2$$
Dividing both sides by $3$ again leaves us with this: 
$$3k^3+1=3t^2$$ 
Which is impossible because $3t^2$ is a multiple of $3$ and $3k^3+1$ is not. 
This is a contradiction. 
Thus the proof is complete. 
</s>