Integer solutions of $a^3+2=3b^3$

Question

I checked this one with my computer and it seemed that there is no solution except $a=b=1$ , but I don't know if it's true or how to prove it. Here is the full question: a, b are natural numbers. Are there any $(a, b)$ other than $(1, 1)$ such that: $a^{3} + 2 = 3b^3$

Answer 1

I don't think my other solution was right so I deleted it, unless someone can prove whether we can assume that $\gcd\left(a^3-1,b^3-1\right)=1$. I hope this solution is correct:

Assume an integer solution $(a,b)$ exists. This means that $$a^3-1=3\left(b^3-1\right)$$Since the cubic residues $\pmod9$ are $0$, $1$ and $8$, so $a^3\equiv1\pmod9\Rightarrow 3\mid\left(a^3-1\right)$. This means that $9\mid\left(a^3-1\right)$. This occurs when $a=3c+1$, where $c\in\mathbb{Z}$. Therefore $$(3c+1)^3-1=9c\left(3c^2+3c+1\right)=3\left(b^3-1\right)$$Similarly, we must have that $3\mid\left(b^3-1\right)$, so $b^3\equiv1\pmod9$ and so $b=3d+1$ where $d\in\mathbb{Z}$. Now we have that $$9c\left(3c^2+3c+1\right)=27d\left(3d^2+3d+1\right)\Rightarrow c\left(3c^2+3c+1\right)=3d\left(3d^2+3d+1\right)$$Since $3\nmid\left(3c^2+3c+1\right)$, we must have that $c=3e$ where $c\in\mathbb{Z}$, so $$27e^2+9e+1=3d^2+3d+1\Rightarrow36e^2+12e+1=4d^2+4d+1\Rightarrow(6e+1)^2=(2d+1)^2$$Square rooting both sides gives $2d+1=\pm(6e+1)$, so $d=3e=c$ or $d=-3e-1=-c-1$.

If $c=d$ then $a=b$, which is the solution $a=b=1$.

If $d=-c-1$ then $3d=-3c-3$ so $b-1=-a-2$ and so $a^3+2=3\left(-a-3\right)^3$ which clearly does not have any integer solutions.

Therefore the only solution is $a=b=1$.