If $x,y,z \in \mathbb{Z}$ such that $x^4+y^4+z^4 \equiv 0 \pmod{29}$, prove that $x^4+y^4+z^4 \equiv 0 \pmod{29^4}$ 
If $x,y,z \in \mathbb{Z}$ such that $x^4+y^4+z^4 \equiv 0 \pmod{29}$, prove that $x^4+y^4+z^4 \equiv 0 \pmod{29^4}$.

I have no idea where to start, but this is my abstract algebra homework, so I think we have to use machinery from abstract algebra. Thanks in advance.
 A: Here is a plan. If you could show that the first congruence is possbile only, if all the variables are divisible by $29$, then you would be done, right?
You could actually brute force it. Cyclicity of the group $\mathbb{Z}_{29}^*$ means that $x^4$ takes seven distinct non-zero values modulo $29$, namely the residue classes in the set
$$S=\{1,7,16,20,23,24,25\}.$$ 
This is basically because $4\mid(29-1)$. Anyway, listing all the $8^3$ possibilities, and checking that "all zeros" is the only solution is doable.
The first reduction to the workload is to observe that no fourth power is $\equiv-1\pmod{29}$. This implies that there is no solution with a single variable divisible by $29$, and the rest non-divisible. So it suffices to exclude the possibility of a solution with all $x^4,y^4,z^4\in S$. Still $7^3$ cases remain (taking into account permutatins of the variables actually a lot less).
But $S$ is clearly also a subgroup of $\mathbb{Z}_{29}^*$. If the sum of any three elements of $S$ is zero, say $u+v+w\equiv0\pmod{29}$, we can multiply this congruence by the inverse of $u$, stay inside the subgroup $S$, and find a solution $1+vu^{-1}+wu^{-1}\equiv0\pmod{29}$. This means that w.l.o.g. we can assume that $u=1$. Only $49$ cases remain. 
As $27\notin S$, we can see that $v\neq1\neq w$, and as $14\notin S$, also $v\neq w$. At this point we take into account the symmetry $v\leftrightarrow w$, and there are ${6\choose 2}=15$ cases remaining. The remaining checks involve verifying that if
$u,v\in\{7,16,20,23,24,25\},u<v$, then $u+v\neq-1\pmod{29}$. This is easy because either $u=7$ and $v\neq21$, or $29<u+v<50$.
The conclusion is that the plan worked. For $x^4+y^4+z^4$ to be divisible by $29$ it is necessary that all the variables are divisible by $29$. Therefore the sum of their fourth powers is divisible by $29^4$.

I think it may be possible to make more clever use of the fact that the elements of $S$ are seventh roots of unity. 
A possibility that occurred to me is to prove that for all triples $(u,v,w)$ of distinct elements of $S\simeq C_7$ we can find an element $x\in S$ such that either $x\{u,v,w\}\cap \{u,v,w\}$ has two elements, or the intersection is empty. In neither case is it possible that $u+v+w=0$ (when also $xu+xv+xw=0$): in the former case we have an obvious contradiction, and in the latter case we contradict the fact that the sum of all elements of $S$ is zero, and the two triples cover all but one element of $S$.
Unfortunately this is not true. Any coding theorist should be familiar with the case that inside a cyclic group of order seven the triple $\{1,g,g^3\}$ intersects all its shifts in a singleton. Something else?
Anyway, checking those fifteen cases is probably simpler. If somebody comes up with some other clever trick, I'll give a bounty.
