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Prove that for any distinct positive integers $a$ and $b$ the number $2a(a^2 + 3b^2)$ is not a perfect cube.

Note that $2a(a^2 + 3b^2) = (a+b)^3 + (a-b)^3.$ If $2a(a^2 + 3b^2)$ were a perfect cube, then the equation $x^3 + y^3 = z^3$ would have a solution in positive integers (both when $a > b$ and when $b > a$). However, the latter claim uses a highly nontrivial theorem, namely Fermat's last theorem for the case n = 3. I was wondering if there's a shorter and more elementary solution to this problem (e.g. infinite descent, Fermat's little theorem, the Euler'fermat theorem, etc.)? Maybe one can assume $a$ and $b$ are coprime? Indeed, write $a=da_1, b=db_1$ where $\gcd(a_1,b_1)=1, d=\gcd(a_1,b_1)$ (note that we didn't need to specify $d=\gcd(a_1,b_1)$). Then $2a(a^2 + 3b^2)$ is a perfect cube iff $2a_1 (a_1^2 + 3b_1^2)$ is a perfect cube (since dividing a perfect cube from a perfect cube yields a perfect cube and multiplying two perfect cubes yields a perfect cube). Then it might be useful to consider the gcd of $2a_1$ and $a_1^2 + 3b_1^2$. We may need additional information on whether $a_1$ and $b_1$ are odd.

user3379
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    You do know that you can prove FLT for $n=3$ with infinite descent right... https://math.stackexchange.com/questions/662313/the-equation-x3-y3-z3-has-no-integer-solutions-a-short-proof – TheBestMagician Sep 16 '22 at 03:33
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    You've proved that if there exists $a$ and $b$ such that $2a(a^2+3b^2)$ is a cube, then there exists $x$ and $y$ such that $x^3+y^3$ is a cube. But the converse is also true: let $a=x+y$ and $b=x-y$. So your problem statement is equivalent to the special case of Fermat's Last Theorem with $n=3$. – Michael Hartley Sep 16 '22 at 03:40
  • @TheBestMagician thanks for the link. I guess there isn't a significantly shorter and simpler solution for this problem after all. – user3379 Sep 16 '22 at 17:36

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