Given a prime $\displaystyle p $, show $\displaystyle \displaystyle x^3+py^3+p^2z^3=0 $ has no nontrivial integer solutions.
Proof by infinite descent:
Suppose there axists a solution $\displaystyle 0\ne (x,y,z)$. Then $\displaystyle p\mid x^3$ since $\displaystyle p\mid (p^3+p^2z^3)$.
Therefore we also have $\displaystyle p\mid x \Rightarrow x = ap$.
Thus, after substitution and division by $\displaystyle p $ we get $\displaystyle y^3+pz^3+p^2a^3 = 0$.
An analogous argument shows that $\displaystyle p \mid y$ and then $\displaystyle p\mid z$ and we get
a new solution $\displaystyle (a,b,c) = (\frac{x}{p},\frac{y}{p},\frac{z}{p})$.
We can proceed with this argument infinitely many times, which is of course absurd.
Note: This problem can be generalized to the statement that there are no
solutions $\displaystyle 0\ne (x_1,x_2,...,x_n) \in \mathbb{Z}^n$ to $\displaystyle \sum_{k=1}^{n} p^kx_k^n = 0$
(EDIT) where $\displaystyle p$ is squarefree.
First part, to prove divisibility by the prime $\displaystyle p $ for $\displaystyle x, y,$ and $\displaystyle z $. From $\displaystyle x^3+py^3+p^2z^3=0 $ it's clear that $\displaystyle p|x $ and then $\displaystyle p|x^3 $. Dividing the equation by $\displaystyle p $ gives $\displaystyle x^3/p+y^3+pz^3=0 $ and since $\displaystyle p $ still divides $\displaystyle x^3/p $ it follows that $\displaystyle p|y $, also $\displaystyle p|y^3 $. Dividing again by $\displaystyle p $ gives $\displaystyle x^3/p^2+y^3/p+z^3=0 $, where $\displaystyle x^3/p^2 $ and $\displaystyle y^3/p $ are divisible by $\displaystyle p $ so $\displaystyle p|z $.
$\displaystyle x^3, y^3,$ and $\displaystyle z^3 $ are of the forms $\displaystyle Ap^3, Bp^3,$ and $\displaystyle Cp^3 $, respectively.Also, $\displaystyle A, B,$and $\displaystyle C $ are relatively prime to $\displaystyle p $. Therefore, $\displaystyle p^{3s}A+p^{3t+1}B+p^{3u+2}C=0 $.
$\displaystyle p^{3s}\neq p^{3t+1}\neq p^{3u+2} $. If $\displaystyle p^{3s} $ is the smallest of these, then after division we have $\displaystyle A+p^{3T+1}B+p^{3U+2}C=0 $. But then $\displaystyle p|A $; since $\displaystyle (p,A)=1 $ and $\displaystyle A>1 $ it's contradiction.
Similarly for the cases where $\displaystyle p^{3t+1} $ or $\displaystyle p^{3u+2} $ are the smallest powers.