Proving $3x^{3}+4y^{3}+5z^{3}\equiv 0 (3^{m})$ has nontrivial integer solutions $\forall$ $m\in\mathbb{N}$ I have to prove that $3x^{3}+4y^{3}+5z^{3}\equiv 0 (3^{m})$ has nontrivial integer solutions $\forall$ $m\in\mathbb{N}$ so that:
$x_{m+1}\equiv x_{m}(3^{m}) $ $\forall$ $m\in\mathbb{N}$ 
$y_{m+1}\equiv y_{m}(3^{m}) $ $\forall$ $m\in\mathbb{N}$ 
$y_{m+1}\equiv y_{m}(3^{m}) $ $\forall$ $m\in\mathbb{N}$ 
where $(x_{m},y_{m},z_{m})$ is a nontrivial solution of $3x^{3}+4y^{3}+5z^{3}\equiv 0 (3^{m})$. I am going to procede like I have done it for each prime $p\neq 3$:
Let $x_{m+1}=x_{m}+a3^{m}$, $y_{m+1}=y_{m}+b3^{m}$, $z_{m+1}=z_{m}+c3^{m}$ for some $a, b, c\in\mathbb{Z}$. Then:
$3x_{m+1}^{3}+4y_{m+1}^{3}+5z_{m+1}^{3}=3(x_{m}+a3^{m})^{3}+4(y_{m}+b3^{m})^{3}+5(z_{m}+c3^{m})^{3}\equiv3x_{m}^{3}+4y_{m}^{3}+5z_{m}^{3}(3^{m+1})$
So, to prove $(x_{m+1}, y_{m+1}, z_{m+1})$ is a solution of that equation for $m+1$, I had to show $(x_{m}, y_{m}, z_{m})$ is a solution for $m+1$. I do not think this is the right way, but I do not get to find any other. Any help would be apreciated. 
 A: To elaborate on the comments : there is no real difficulty here. Let
$f(x,y,z)=3x^3+4y^3+5z^3$. One can
construct by induction a sequence $(x_m,y_m,z_m,t_m)_{m\geq 1}$ 
of elements of ${\mathbb N}^4$ such that for all $m\geq 1$,
$x_{m+1}\equiv x_m({\sf mod} \ 3),y_{m+1}\equiv y_m({\sf mod} \ 3),
z_{m+1}\equiv z_m({\sf mod} \ 3)$ and
$$
f(x_m,y_m,z_m)=3^{m+1}t_m \tag{1}
$$
Notice the $3^{m+1}$ (not $3^m$) in the LHS. This is probably the 
stumbling block that prevented you from finding the solution.
We have
$$
f(x_m+3^ma,y_m+3^mb,z_m+3^mc) \equiv 
3^{m+1}(t_m+4by_m^2+5cz_m^2) \ ({\sf mod}\ 3^{m+2}) \tag{2}
$$
We wish to find $b,c$ such that $t_m+4by_m^2+5cz_m^2$ is divisible by $3$,
and this is always possible when one of $y_m$ or $z_m$ is not  divisible by $3$.
From (2) it is now clear how to proceed by induction : we must 
 change the induction hypothesis slightly, by replacing (1) with
$$
f(x_m,y_m,z_m)=3^{m+1}t_m, y_m\not\equiv 0({\sf mod}\ 3)
\text{ or } z_m\not\equiv 0({\sf mod}\ 3)\tag{1'}
$$
 This concludes the construction (don't forget to set an initial value for the sequence ...).
