Quite Standard. The pith of this problem is the following three-variable version of Thue's theorem: Claim. [Three-Variable Thue] Let $p$ be any prime number and let $a,b,c$ be positive integers at most $p-1$. Then, there exist three integers $x,y,z$ not all zero, whose absolute values is at most $\sqrt[3]{p}$ each for which $p\mid ax+by+cz$. Proof. We do the same thing as in the proof of Thue, run the residues of $ax+by+cz$ modulo $p$ for all triples $(x,y,z)$ for which $x,y,z$ range from $0$ to $\lfloor\sqrt[3]{p}\rfloor$ each. Since the number of such triples exceeds $p$, we see that by the pigeon hole principle, some two of the triples $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ exist whose values are equivalent modulo $p$, so choosing $(x,y,z) = (x_1-x_2,y_1-y_2,z_1-z_2)$ finishes and it is clear that each of these values in the triple satisfy the condition that their absolute values are at most $\sqrt[3]{p}$. Back to the main problem, we see that the expression given in the problem nicely factorizes, due to Titu's factorization: $$x^3+2y^3+4z^3-6xyz = \left(x+\sqrt[3]{2}y+\sqrt[3]{4}z\right)\left[x^2+(\sqrt[3]{2}y)^2+(\sqrt[3]{4}z)^2-x\sqrt[3]{2}y-\sqrt[3]{2}y\sqrt[3]{4}z-\sqrt[3]{4}zx\right]$$ Now we are clear on what to do: Note that the map $x\rightarrow x^3$ is surjective modulo $p$ where $p\equiv 2\pmod{3}$, so we now center our focus on only such primes, and fix such a prime $p$. Let $\epsilon$ be the cube root of $2$ modulo $p$, that is, $\epsilon^3\equiv 2\pmod{p}$, which obviously exists due to the surjectivity of the map as described earlier. Then, from the above claim, let $u,v,w$ be integers whose absolute value is at most $\sqrt[3]{p}$ for which $p\mid u+\epsilon v+\epsilon^2 w$, then we see that for such a prime $p$, due to the great factorization above, we see that: $p\mid u^3+2v^3+4w^3-6uvw$. Moreover, the key part which we should be noting is, due to the bound on $u,v,w$, we have that: $$\lvert u^3+2v^3+4w^3-6uvw\rvert < 13p$$ and the second important part is that we can always make sure that the given quantity is nonnegative. Indeed, if we have that the given quantity is negative, replace $(u,v,w)$ with $(-u,-v,-w)$. With this framework in mind, we can finally slander the problem: For each prime $p\equiv2\pmod{3}$, define $u_p,v_p,w_p,\theta_p$ to be integers (the last one nonnegative) so that $$u_p^3+2v_p^3+4w_p^3-6u_pv_pw_p=p\theta_p$$ where $\theta_p$ is a nonnegative integer strictly less than $13$. So, as we vary $p$ over all primes of the form $3k+2$, we see that this $\theta_p$ must be common for infinitely many primes $p$ taking that form, denote that common value by $\theta$. If $\theta$ is non-zero, then, $n=\theta$ is our desired positive integer as requested in the problem, otherwise, it means that the equation $x+\sqrt[3]{2}y+\sqrt[3]{4}z=0$ or $x=\sqrt[3]{2}y=\sqrt[3]{4}z$ has infinitely many solutions over integers, so that no two of them scale each other, impossible. Therefore $\theta$ is indeed positive and so we are done. Remark. In my opinion, such problems become really intuitive (not easy) when one is familiar with the understanding of what 'standard nt' (not easy nt) really is. Another problem whose general framework is isomorphic to that of this problem is Euler-Lagrange : Quote: Prove that if $p$ is a prime of the form $1,9\pmod{20}$, then positive integers $a,b$ exist for which $p=a^2+5b^2$ and if $p\equiv 3,7\pmod{20}$, then we have instead $2p=a^2+5^2$.

Conclusively I feel this is a nice problem. If you're interested in a similar contest problem, try to solve: Spoiler hehe

An alternative junior-ish way to argue $a^3 + 2b^3 + 4c^3 - 6abc \neq 0$ is to do an infinite descent mod $2$. Here is a Algebraic Number Theory styled solution. (If not mistaken, it actually follows that $p$ itself is of the desired form, correct me if I am wrong). Note that $p\equiv 2 \pmod 3$ implies that $a^3 = 2 \pmod p$ for some integer $a$, this implies that $p$ is not a prime in $\mathbb{Z}[2^{1/3}]$ (otherwise $p$ divides $a - 2^{1/3}$ or $a^2 + a \cdot 2^{1/3} + 2^{2/3}$, impossible by comparing $2^{1/3}$-parts), hence $p$ is reducible in $\mathbb{Z}[2^{1/3}]$ (since $\mathbb{Z}[2^{1/3}]$ has class number $1$, can be computed in the standard Minkowski bound approach, hence a UFD, so irreducibles and primes are the same), i.e. $p = xy$ for some $x,y \in \mathbb{Z}[2^{1/3}]$ of norm not equal to $1$. Thus $p^3 = N(x)N(y)$ and at least one of the multipliers on the right is equal to $\pm p$, hence the result due to $N(a + 2^{1/3}b + 2^{2/3}c) = a^3 + 2b^3 + 4c^3 - 6abc$ (if equal to $-p$, then just replace $a,b,c$ with their additive inverses).

Both proofs are very nice! I just want to add that the expression is the same as USATST 2012. Both problems can be solved using the norm approach.