<s>[INST] Show that there exist no $a, b, c \in \mathbb Z^+$ such that $a^3 + 2b^3 = 4c^3$ 
Find all positive integer solutions of $a^3 + 2b^3 = 4c^3$.

Proof: There don't exist any integer solutions for the give equation.
Proof by the Well Ordering Principle.
Let $d$ be the set of all such possible combinations in form $(a, b, c)$.
By the Well Ordering Principle, we that every nonempty set of non-negative integers always has a smallest element. So, there must be a tuple $(a, b, c)$ such that $\max(a, b, c)$ is the least possible value in $d$.
We also know that $a$, $b$ and $c$ must be even. Since $a$ must be even if $4c^3$ has to be even but we know that $4c^3$ is even, so $a$ is the same. Similarly it can be shown that all $a$, $b$ and $c$ are even.
Now, let $a = 2x$, $b = 2y$ and $c = 2z$.
But then, 
$$a^3 + 2b^3 = 4c^3$$
$$\implies (2x)^3 + 2*(2y)^3 = 4*(2z)^3$$
$$\implies 8x^3 + 16y^3 = 32z^3$$
$$\implies x^3 + 2y^3 = 4z^3$$
So, $(x, y, z) \in d$. But, $\max(x, y, z) < \max(a, b, c)$. But $(a, b, c)$ was supposed to be the smallest such member. We have a contradiction here. There is no smallest $(a, b, c)$ which satisfy this criterion and hence no $(a, b, c)$ at all and $d$ is empty. Hence, there exist no integers $a$, $b$ and $c$ which satisfy  $a^3 + 2b^3 = 4c^3$
I wrote this proof. Is it enough to prove the above question? Am I right? Are there any mistakes in the above proof? If there are any, please help me. 
Do you have any other ways of proving it?
 [/INST] There is another proof using the well ordering principle.
We use the following fact:

Third powers are can only be congruent to $0 $,  $1 $ or  $8 \mod 9 $. 

Now, we see that the left hand side is congruent to $0$, $1$, $2$, $3$, $6$, $7$ or $8 \mod 9$, while the right hand side is congruent to $0, 4$ or $5 \mod 9$. So the only possibility is that $4c^3 \equiv 0 \mod 9$ and $a^3+2b^3 \equiv 0 \mod 9$. The first modular congruence gives that $3 \mid c$. The second can only hold true if $3 \mid a$ and $3 \mid b$. 
If one of those isn't true, then we get either a combination of $0 \mod 9$ and $2$ or $7 \mod 9$, a combination of $0 \mod 9$ and $1$ or $8 \mod 9$, a combination of $1$ or $8 \mod 9$ and $2$ or $7 \mod 9$, none of which add to $0 \mod 9$. 
Now suppose that $(a,b,c)$ is the solution with the smallest $a$. 
We know that $3 \mid c$, $3 \mid a$ and $3 \mid b$. 
Hence write $a=3x, b=3y, c=3z$ and then is $(x,y,z)$ also a solution with smaller $a$. 
</s>