Simple Diophantine Equation

**chiph588@** · August 16th 2010, 02:08 PM

Given a prime $p$ , show $\displaystyle x^3+py^3+p^2z^3=0$ has no nontrivial integer solutions.

**Unbeatable0** · August 16th 2010, 02:21 PM

Proof by infinite descent:

Suppose there axists a solution $0\ne (x,y,z)$ . Then $p\mid x^3$ since $p\mid (p^3+p^2z^3)$ .

Therefore we also have $p\mid x \Rightarrow x = ap$ .

Thus, after substitution and division by $p$ we get $y^3+pz^3+p^2a^3 = 0$ .

An analogous argument shows that $p \mid y$ and then $p\mid z$ and we get

a new solution $(a,b,c) = (\frac{x}{p},\frac{y}{p},\frac{z}{p})$ .

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 $0\ne (x_1,x_2,...,x_n) \in \mathbb{Z}^n$ to $\sum_{k=1}^{n} p^kx_k^n = 0$

(EDIT) where $p$ is squarefree.

**melese** · August 17th 2010, 12:30 PM

First part, to prove divisibility by the prime $p$ for $x, y,$ and $z$ . From $x^3+py^3+p^2z^3=0$ it's clear that $p|x$ and then $p|x^3$ . Dividing the equation by $p$ gives $x^3/p+y^3+pz^3=0$ and since $p$ still divides $x^3/p$ it follows that $p|y$ , also $p|y^3$ . Dividing again by $p$ gives $x^3/p^2+y^3/p+z^3=0$ , where $x^3/p^2$ and $y^3/p$ are divisible by $p$ so $p|z$ .

$x^3, y^3,$ and $z^3$ are of the forms $Ap^3, Bp^3,$ and $Cp^3$ , respectively.Also, $A, B,$ and $C$ are relatively prime to $p$ . Therefore, $p^{3s}A+p^{3t+1}B+p^{3u+2}C=0$ .

$p^{3s}\neq p^{3t+1}\neq p^{3u+2}$ . If $p^{3s}$ is the smallest of these, then after division we have $A+p^{3T+1}B+p^{3U+2}C=0$ . But then $p|A$ ; since $(p,A)=1$ and $A>1$ it's contradiction.
Similarly for the cases where $p^{3t+1}$ or $p^{3u+2}$ are the smallest powers.

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