Solve over natural numbers: $m^3=2n^3+6n^2$. Functional equation gives rise to a diophantine equation! My question is basically to find all natural numbers $(m,n)$ such that
$m^3=2n^3+6n^2$
First for some background (this is not really that relevant but anyways):
I was trying to solve an olympiad functional equation
$f:\mathbb{N} \to \mathbb{N}$ such that $f(x+y)=f(x)+f(y)+3(x+y)\sqrt[3]{f(x)f(y)}$
Then we will get that $f(x)$ will always be a perfect cube so setting $f(x)=g(x)^3$ and then $P(1,1)$ gives $g(2)^3=2g(1)^3+6g(1)^2$. And then I noticed that if $g(1)=1$ we could just induct up to get $g(x)=x$ always, so this is where the question comes from, because if i prove the only solution in natural numbers to the diophantine in $(1,2)$ I will be done.
Now back to the diophantine, obvioulsy $(m,n)=(2,1)$ is a solution and wolfram alhpha says it is the only solution. For the past one hour or so I have been trying to prove this but have done basically nothing...
I have just been doing random subsitutions and making cases and stuff, like $m=2y$ so then we get $y \equiv x \mod 3$
and then making cases like if $x$ is divisible by $3$ or not but I have not even been able to rule out one case...
There are infact more integer solutions so I believe we could use some inequalities somehow but I am not sure how whatsoever...
Any help on this problem will be appreciated...
Thanks!
 A: This is not a complete solution but it is too long for a comment.
Let $g=(m,n)$ and $m=ag,n=bg$ with $(a,b)=1$. Then $a^3g=2b^2(bg+3)$ and so $b^2\mid g$, giving $g=kb^2$, transforming the equation eventually to
$$k(a^3-2b^3)=6.$$ This gives four equations $a^3-2b^3=c$ where $c\in \{1,2,3,6\}$.
Cases $a^3-2b^3=1$ and $a^3-2b^3=2$ can be both resolved using Skolem's theorem (see also this answer ):

For any integer $d \ne 0$, there exists at most one pair $(x,y) \in \mathbb{Z} \times \mathbb{Z}$ with $y \ne 0$ such that $$x^3 + dy^3 = 1$$

We can see that for $d=2$ such non-trivial solution is $(x,y)=(-1,1)$ and for $d=4$ the linked resource shows there is no non-trivial solution.
For $a^3-2b^3=1$ using the above (and writing $a^3+2(-b)^3=1$) we get only solutions $(-1,-1)$ and $(1,0)$. For $a^3-2b^3=2$ we let $a=2u$ and the equation translates to $4u^3-b^3=1$, which fits the above theorem as $(-b)^3+4u^3=1$. So using the above result again, we have $(b,u)=(-1,0)$ as the only solution. None of these solutions correspond to a positive solution of the original problem.
So it "only" remains to solve $a^3-2b^3=3$ and $a^3-2b^3=6$ (which can be transformed to $a^3-4b^3=3$ using that $a$ must be even). Note that former has solutions $(-5,-4)$ and $(1,-1)$ which are not positive, while the latter has a solution $(a,b)=(2,1)$, which corresponds to the solution noted in the OP. Whether these are the only solutions, I do not know.
A: This doesn't answer the diophantine equation $ m^3 = 2n^3 + 6n^2$ (which is hard), but answers the functional equation $f(x+y) = f(x) + f(y) + 3(x+y) \sqrt[3]{f(x) f(y) } $ when defined over all integers $ f: \mathbb{Z} \rightarrow \mathbb{Z}$.
With $ x = y = 0$,  $f(0) = 0$.
Now, suppose there is some $ a \neq 0 $ such that $f(a) = 0$, then with $ y = a$, we get that $f(x+a) = f(x)$ , so we have a periodic function. If there is some $x$ with $f(x) \neq 0$, then $f(x+x+a) = f(x+x)$ gives us $ 3(x+x+a)\sqrt[3]{f(x)f(x) } = 3(x+x)\sqrt[3]{f(x) f(y)}$, which is a contradiction. Hence, the only solution is  $f(x) = 0 \, \forall x$ (which doesn't give us a solution on the positive integers).
Otherwise, $f(x) \neq 0 \Leftrightarrow x \neq 0$.
We will show that $\forall x, y \neq 0$,
$$\frac{f(x) } { x^3 } = \frac{f(y) } { y^3}.$$
This is obvious if $ x = - y$ by definition.
Otherwise, let $ z = - x - y \neq 0$. So $-f(x) = f(y+z) = f(y) + f(z)  - 3x \sqrt[3]{f(y) f(z) }$, or that $ f(x) + f(y) + f(z) = 3x \sqrt[3]{ f(y) f(z) } $.
Swapping $x$ and $y$, we get that $ f(y) + f(x) + f(z) = 3y \sqrt[3]{f(x) f(z) }$.
Hence, $ \frac{f(x) } { x^3 } = \frac{f(y) } { y^3}$ as claimed. This has value $f(1)$.
Finally, setting $f(x) = f(1)x^3$ with $f(1) \neq 0$ in the functional equation, we get
$$f(1) ( x+y)^3 = f(1) x^3 + f(1) y^3 + 3(x+y) f(1)^{2/3} xy$$
or that $3xy(x+y) ( f(1) - f(1) ^ { 2/3} ) = 0 $.
This gives us $ f(1) = 1$ so $ f(x) = x^3 \, \forall x \in \mathbb{N}$.
Note

*

*Instead of solving $ m^3 = 2n^3 + 6n^2$, we're solving $ 8n^3 = 2n^3 + 6 n^2$ which is much easier.

*We might be able to work around extending the function to the negative integers (EG replacing $f(z) = f(-x-y) = - f(x+y)$), but that solution presentation felt the most natural to me.

*In fact, this solution also works if the domain and range are the reals. We didn't use any facts about integers / number theory.


Ignore this:
Suppose that the function is only defined for the positive integers.
We extend the function to the non-negative integers by defining $f(0) = 0$.
It remains to check that for $ x, y \geq 0$, we still have $ f(x+y) = f(x) + f(y) + 3(x+y) \sqrt[3]{f(x)f(y)}$, which is obvious in the cases of $ x = 0 , y = 0 , x = y = 0$.
Again, we extend the function to the negative integers by defining $f(-x) = - f(x)$. We have to check that the functional equation holds, which is left as an exercise to the reader.
Of course, we now have a superset of solutions, and have to check which allow for solutions to the positive integers.
