3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 '17 at 20:48
  • 1
    $\begingroup$ @DietrichBurde We could have $x_1 = 1,x_2 = 1+(\sqrt[3]{3}-1)$. $\endgroup$ – user19405892 Jan 24 '17 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 '17 at 21:07
  • $\begingroup$ @dietrich. You also have to show $d$ is an integer. $\endgroup$ – fleablood Jan 24 '17 at 21:19
  • 5
    $\begingroup$ Not fair to change the question! $\endgroup$ – Robert Israel Jan 24 '17 at 21:36
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Um $x_3^3 = -t^2$ is not a perfect square. $\endgroup$ – fleablood Jan 24 '17 at 21:26
  • 1
    $\begingroup$ $x_1^3 = t^2$, $x_1^3 + x_2^3 = t^2$, $x_1^3 + x_2^3 + x_3^3 = 0$ are all squares. $\endgroup$ – Robert Israel Jan 24 '17 at 21:27
  • 1
    $\begingroup$ Oops You are right. $\endgroup$ – fleablood Jan 24 '17 at 21:27
  • $\begingroup$ I changed the question to four terms. $\endgroup$ – user19405892 Jan 24 '17 at 21:30
  • 1
    $\begingroup$ 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 '17 at 21:41
4
$\begingroup$

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} $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.