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Find all pairs of integers $(x,m)$ for which $$\sqrt[3]{\sqrt[3]{x-2}+m}+\sqrt[3]{-\sqrt[3]{x-2}+m}=2$$ hold.

I have used this property :

Property: if $$a+b+c=0 \implies a^3+b^3+c^3=3abc, $$ I come up to the following equation: $(2m-8)^3=-216(m^2-(x-2)^{2/3})$ , such that $a=-2, b=(\sqrt[3]{x-2}+m)^{1/3}, c=(-\sqrt[3]{x-2}+m)^{1/3}$, I can't solve the last equation however i tried $x$ as a paramater instead of $m$ , The solution from wolfram alpha are $(x,m)=(2,1),(66,4)$, Any Help ?

nonuser
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zeraoulia rafik
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  • For $x,m$ integers, LHS is integer (equal to $2$) or algebraic of $9$ or $3$ degree. It follows LHS must be integer so $x-2,\space \sqrt[3]{x-2}+m$ and $- \sqrt[3]{x-2}+m$ need to be cubes. Since $\sqrt[3]{(t+1)^3}+\sqrt[3]{(-t+1)^3}=2$ and $$(\pm t+1)^3=\pm t^3+3t^2+\pm3t+1=\pm(t^3+3t)+(3t^2+1)$$ we get a parametrization of infinitely many solutions $$x-2=(t^3+3t)^3\m=3t^2+1$$ – Piquito Aug 23 '21 at 22:29
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    I would have expected $\sqrt[3]{\sqrt{x-2}+m}+\sqrt[3]{-\sqrt{x-2}+m}=2 ; ; ; $ – Will Jagy Aug 24 '21 at 01:04
  • @WillJagy I think just adding some modifications about it it will be hold – zeraoulia rafik Aug 24 '21 at 17:33
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    The reason to expect my revised form is Cardano's method for the roots of a cubic. https://en.wikipedia.org/wiki/Cubic_equation#Cardano's_formula – Will Jagy Aug 24 '21 at 17:38

2 Answers2

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Let $y=\sqrt[3]{x-2}$, then we have $$2 = \sqrt[3]{m+y} + \sqrt[3]{m-y}\;\;\;\;\;\;|^3$$ $$8 = m+y +3\sqrt[3]{m^2-y^2}\cdot 2 +m-y$$ and thus $$4-m = 3\sqrt[3]{m^2-y^2} $$ so $$ 64 -48m+12m^2-m^3 = 27m^2-27y^2$$ or $$27y^2 =m^3+15m^2+48m-64$$ and finally $$\boxed{27y^2 = (m-1)(m+8)^2}$$ Since $\gcd(m-1,m+8) \mid 9$ we have $3\mid m+8$ and $3\mid m-1$ and so $m+8 = 3b$ and $m-1 = 3a$ so $$y^2 = ab^2\implies a=c^2$$ for some $c\in\mathbb{N}$.

Now $\boxed{m = 3c^2+1}$ and $ y= bc = c(c^2+3)$ so $\boxed{x= 2+c^3(c^2+3)^3}$

  • For $c=0$ we get $m=1$ and $x= 2$
  • For $c=1$ we get $m=4$ and $x= 66$
  • For $c=2$ we get $m=13$ and $x= 14^3+2$
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nonuser
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Generally solutions $(x,m)$ can be split into three cases:

  • $(x,m)=\left(\frac{1}{243} \left(486-\sqrt{3} (m-1)^{3/2} (m+8)^3\right),m\right)$
  • $(x,m)=\left(\frac{1}{243} \left(486+\sqrt{3} (m+8)^3 (m-1)^{3/2}\right)),m\right)$
  • $(x,m)=(2,1)$

In the first two cases $m\ge2$ and we choose $m$ such that the resulting term for $x$ is an integer. For example if we choose $m=4$, the first case yields the solution $(x,m)=(-62,4)$ and from the second case we obtain $(x,m)=(66,4)$.

You can generate infinitely such solution pairs by choosing a $m$ where $m-1$ is an odd cubic, e.g. $m=3^1+1,3^3+1,3^5+1,\ldots$ or, more precisely, as shown by @Aqua where $m=3c^2+1$ for natural $c$.