# Simple Diophantine Equation

• Aug 16th 2010, 02:08 PM
chiph588@
Simple Diophantine Equation
Given a prime $\displaystyle p$, show $\displaystyle \displaystyle x^3+py^3+p^2z^3=0$ has no nontrivial integer solutions.
• Aug 16th 2010, 02:21 PM
Unbeatable0
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.
• Aug 17th 2010, 12:30 PM
melese
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.