How prove this $ n = \frac {a^3 + 2b^3} {c^3 + 2d^3}. $ infinitely many special numbers Qustion

A  special number  is a positive integer $n$ for which there exists positive integers $a$, $b$, $c$, and $d$ with $$ n = \dfrac {a^3 + 2b^3} {c^3 + 2d^3}. $$ Prove that

i) there are infinitely many special numbers;
ii) $2014$ is not a special number.
I have solve (1):let $$b=c=d=1,a=3k+1$$ then
$$n=9k^3+9k^2+3k+1$$ is sepecial for any postive integer $k$,Clearly,these numbers are infinitely many.
But for (2),Assmue that 
$$2014(c^3+2d^3)=a^3+2b^3\Longrightarrow  a^3+2b^3=2\cdot 19\cdot 53(c^3+2d^3)$$
then I can't prove contradiction.I guess can use (1) to solve (2).But I can't. Thank you

First note that $-2$ is not a cubic residue modulo $19$. To see this, suppose
$x^3\equiv -2\pmod{19}$. Then $x^{18}\equiv (-2)^6 \equiv 64 \equiv 7\pmod{19}$
Clearly $19$ does not divide $x$, or we would have $x^3\equiv 0\pmod{19}$, and so by Fermat's Little Theorem we obtain $x^{18}\equiv 1\pmod{19}$, which is a contradiction.
Now suppose that $19 \mid (x^3+2y^3)$. If $19$ divides either $x$ or $y$ then it must divide both. Suppose $19$ divides neither $x$ nor $y$ and let $z$ be the multiplicative inverse of $y$ modulo $19$. We then have
$x^3\equiv -2y^3\pmod{19}$ and so $(xz)^3\equiv -2\pmod{19}$ which is a contradiction.
We see that if $19\mid(x^3+2y^3)$ for some $x$ and $y$ then it follows that $19\mid x$ and $19\mid y$.
Now returning to the problem:
Suppose that $(a^3+2b^3)=2014(c^3+2d^3)$, and - furthermore - suppose that this quadruple $(a,b,c,d)$ is the smallest in the sense that $a$ has the minimum possible value for any quadruple $(a,b,c,d)$ satisfying this equation. We'll show that there is a solution will a smaller value for a, which is a contradiction.
Since $19\mid 2014$ we see that $19\mid(a^3+2b^3)$ and so $19\mid a$ and $19\mid b$ by our observations above.
Let $a=19a_1$ and $b=19b_1$
Then $19^2(a_1^3+b_1^3)=106(c^3+2d^3)$. We see that $19\mid(c^3+2d^3)$ and so we also have $19\mid c$ and $19\mid d$. Now let $c=19c_1$ and $d=19d_1$. We then obtain
$a_1^3+2b_1^3=2014(c_1^3+2d_1^3)$. This gives us a solution to the equation with a smaller value for $a$ (namely $a_1=\frac{a}{19}$), which is a contradiction.
