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This result has been disproven.

Let $x_1,x_2,x_3,x_4$ be an arithmetic progression and suppose that $$x_1^3, \quad x_1^3 + x_2^3, \quad x_1^3 + x_2^3 + x_3^3, \quad x_1^3 + x_2^3 + x_3^3+x_4^4$$ are perfect squares. Prove that $x_1,x_2,x_3,x_4$ are all integers.

Since these numbers are in arithmetic progression, we must have $x_1,x_1+d,x_1+2d,x_1+3d$ are our four terms. Also, $x_1$ must either be a perfect square, or the cubic root of a perfect square and $x_2,x_3$ must be integers or cubic roots. I thought about proving the latter is impossible by showing that the distance between the cubic roots of integers is unique ($\sqrt[3]{a}-\sqrt[3]b)$, so that $x_4^3$ can't be an integer. How do we continue from here?

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  • $\begingroup$ We only have to show that $x_1$ is an integer, because then $x_2=x_1+d$ and $x_3=x_1+2d$ are integers as well, right? $\endgroup$
    – Dietrich Burde
    Jan 24, 2017 at 20:48
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    $\begingroup$ @DietrichBurde We could have $x_1 = 1,x_2 = 1+(\sqrt[3]{3}-1)$. $\endgroup$
    – user19405892
    Jan 24, 2017 at 20:50
  • $\begingroup$ Ah, so $d$ need not be an integer, as it is often the definition for arithmetic progressions. $\endgroup$
    – Dietrich Burde
    Jan 24, 2017 at 21:07
  • $\begingroup$ @dietrich. You also have to show $d$ is an integer. $\endgroup$
    – fleablood
    Jan 24, 2017 at 21:19
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    $\begingroup$ Not fair to change the question! $\endgroup$
    – Robert Israel
    Jan 24, 2017 at 21:36

2 Answers 2

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Not true. Let $x_1 = t^{2/3}$, $x_2 = 0$, $x_3 = -t^{2/3}$ for any positive integer $t$ that is not a cube.

EDIT: For the new question, it's still not true (assuming there are nontrivial integer solutions). Given any integer solution $(x_1, x_2, x_3, x_4)$, not all $0$, multiply all of these by $2^{2/3}$ and you get a non-integer solution.

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  • $\begingroup$ Um $x_3^3 = -t^2$ is not a perfect square. $\endgroup$
    – fleablood
    Jan 24, 2017 at 21:26
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    $\begingroup$ $x_1^3 = t^2$, $x_1^3 + x_2^3 = t^2$, $x_1^3 + x_2^3 + x_3^3 = 0$ are all squares. $\endgroup$
    – Robert Israel
    Jan 24, 2017 at 21:27
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    $\begingroup$ Oops You are right. $\endgroup$
    – fleablood
    Jan 24, 2017 at 21:27
  • $\begingroup$ I changed the question to four terms. $\endgroup$
    – user19405892
    Jan 24, 2017 at 21:30
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    $\begingroup$ Well, the new question may be true if the only integer solution is $(0,0,0,0)$. @user19405892, Robert gives you a counterexample, provided that there exists a nonzero integral solution $\left(x_1,x_2,x_3,x_4\right)$. $\endgroup$
    – Batominovski
    Jan 24, 2017 at 21:41
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Here's a counter example for every $n$ and for every $t$ which is not a third power of a integer. $$x_n=n\cdot t^{2/3}\\x_1^3+x_2^3+\cdots+x_n^3=t^2(n^3+(n-1)^3+\cdots+1^3)=t^2\left(\frac{n(n+1)}{2}\right)^2$$ For example take $x_1=2^{2/3},x_2=2\cdot2^{2/3},x_3=3\cdot 2^{2/3},x_4=4\cdot 2^{2/3} $

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