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Does there exists any positive integer solution to the equation $x^3+y^3=2z^3$ where $x\neq y\neq z$. I tried to find but could not get such triple. We all know that $x^3+y^3=z^3$ don't have a integer solutions. But here if $x=y=z$ gives us a solution. But whether there exists a solution where integers are distinct.

rubina saikia
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    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Mar 19 '23 at 10:35
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    Folks, unless OP has a good grasp of Algebraic Number Theory, it's unreasonable to expect OP to have much more to say about this question than what's already there. Please let the question stay (unless it's a duplicate, of course). – Gerry Myerson Mar 19 '23 at 10:48
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    See here for a duplicate (see the references in the answer). – Dietrich Burde Mar 19 '23 at 11:25
  • I would try $x^3-z^3=z^3-y^3$, then factor both sides and find the conditions that provide solutions ( or the lack of solutions ). – user25406 Mar 20 '23 at 12:06

1 Answers1

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Theorem 3 of Chapter 15 of Mordell, Diophantine Equations, says,

Let $a=p$ or $p^2$, where $p\equiv2,5\bmod9$ is prime, and let $\epsilon$ be a unit in ${\bf Q}(\rho)$ [Note: $\rho$ is a complex cube root of unity]. Then the equation $$ x^3+y^3+\epsilon az^3=0, $$ has no solutions $(x,y,z,\epsilon)$ in the field ${\bf Q}(\rho)$ except $$ z=0,\qquad x=-y,-\rho y,-\rho^2y, $$ unless $a=2$, when there are also the solutions $$ x^3=y^3=-\epsilon z^3,\qquad\epsilon=\pm1. $$ This applies to the case $a=2$, $\epsilon=-1$, and settles the question.

Gerry Myerson
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