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The question is to show that the equation $$x^3+2y^3+4z^3=0 $$ has no (non trivial) integer solutions. I know it can be done using infinite descent, but how do you do it using only the modular arithmetic argument?

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    $\begingroup$ Hint: modulo 8. $\endgroup$ – Arthur Jan 4 '17 at 15:30
  • $\begingroup$ I'm not sure exactly what to do though? $\endgroup$ – Anon Jan 4 '17 at 15:32
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    $\begingroup$ The answers below show you how, although it's still infinite descent (proving that any given solution has common divisors). It's not possible to do better with modular arithmetic, because $x\equiv y\equiv z\equiv 0$ is a solution. $\endgroup$ – Arthur Jan 4 '17 at 16:06
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HINT: Well take a minimal solution $(x',y',z')$ for which $\gcd(x',y',z')=1$ then looking mod $2,4,8$ conclude first that $2\mid x$ then $2\mid y$ then $2\mid z$ so $\gcd(x',y',z')\geq 2$ hence contradiction.

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Mod $9$ also solves it.

By Binomial Theorem $(3k\pm 1)^3\equiv \pm 1\pmod{9}$,

so $x^3\equiv\{0,\pm 1\}\pmod{9}$

(also $x^3\equiv \{0,\pm 1\}\pmod{8}$, so mod $8$ solves it similarly, as a comment has suggested).

$x^3+2y^3+4z^3\equiv 0\pmod{9}$

is equivalent to $x^3+2y^3\equiv -4z^3\pmod{9}$

$x^3+2y^2\equiv \{\pm 3,\pm 2, \pm 1, 0\}\pmod 9$

while $-4z^3\equiv \{0,\pm 4\}\pmod{9}$.

Therefore $x^3+2y^3\equiv -4z^3\equiv 0\pmod{9}$.

So $3\mid x,y,z$. Therefore $(x,y,z)$ is a solution if and only if $\left(\frac{x}{3},\frac{y}{3},\frac{z}{3}\right)$ is also a solution, so $(x,y,z)=(0,0,0)$ (see, e.g., proof by infinite descent). As a comment has said, you must use infinite descent because $x\equiv y\equiv z\equiv 0$ is a solution mod anything.

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