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Suppose we define $A= (8+\sqrt{x})^{1/3} + (8-\sqrt{x})^{1/3}$. How can we find, algebraically, all values of x for which $A$ is an integer?

I was not able this problem save for with Mathematica. How can we solve this using the tool of our brains?

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$A^3 = 16+3A \sqrt[3]{64-x}$, $A^3 = 16+3A y$. –  Oleg567 Nov 10 '12 at 2:45
    
for each $A \ne 0$ we have $y=\frac{A^3-16}{3A}$, and $x=64-y^3$. –  Oleg567 Nov 10 '12 at 2:51

1 Answer 1

$A^3 = 16+3A\sqrt[3]{64-x}$, or $A^3 = 16+3Ay$, where $y=\sqrt[3]{64-x}$.

For each $A \ne 0$ we have $y=\frac{A^3-16}{3A}$, and $x=64-y^3$.

If $A\in (0,4]$, then $x\ge 0$.
If $A\in (4,\infty)$, then $x<0$, so $\sqrt{x}$ will be imaginary.
(if $A<0$, then we have no solutions, because RHS of your formula has $Re > 0$. (?))

Writing by one row, we have $$x=64-\left(\frac{A^3-16}{3A}\right)^3.$$

\begin{array}{|l|l|} A & x \\ --- & --- \\ 1 & 189 \\ 2 & 1792/3^3 \\ 3 & 45325/3^6 \\ 4 & 0 \\ 5 & -1079029/(3\cdot 5)^3 \\ 6 & -7626752/(3\cdot 6)^3 \\ 7 & -34373079/(3\cdot 7)^3 \\ 8 & -236600/3^3 \\ 9 & -361207385/(3\cdot 9)^3 \\ \cdots & \cdots \end{array}

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Works for me. The mistake I made was assuming (without realizing any assumption was being made, a standard mind disassociation) that my variable $p= - 3 y$ compared with your $y,$ was an integer. I imagine my answers with $A<0$ amount to non-principal cube roots, as in Cardano's method. Anyway, +1. –  Will Jagy Nov 10 '12 at 21:13
    
Thank you very much :) –  Oleg567 Nov 10 '12 at 21:17

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