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So the homework question goes as follows:

Find rational numbers such that $ \sqrt[3]{7+5\sqrt{2}}=\alpha+\beta\sqrt{2}$

The homework only asks for one pair of $\alpha$ and $\beta$ and it is almost trivial to find the solution $\alpha=\beta = 1$

Solution:

$\sqrt[3]{7+5\sqrt{2}}=\alpha+\beta\sqrt{2}$

$7+5\sqrt{2}=(\alpha+\beta\sqrt{2})^3$

$7+5\sqrt{2}=\alpha^3+6\beta^2\alpha +\sqrt{2}(3 \alpha^2 \beta +2\beta^3)$

By identification of coefficients:

$7=\alpha^3+6\beta^2\alpha \space and \space 5=3 \alpha^2 \beta +2\beta^3$

So I just "saw" that $\alpha=\beta = 1$ is a solution

My question is, how do we find other solutions?

Thank you!

Rick Decker
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wwbb90
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  • If you have a possibly different solution, $z = \alpha + \beta\sqrt{2}$ with $(\alpha,\beta) \neq (1,1)$, then $\left(\frac{\alpha+\beta\sqrt{2}}{1+\sqrt{2}}\right)^3 = \left((2\beta-\alpha) + (\alpha-\beta)\sqrt{2}\right)^3 = 1$. – Daniel Fischer Sep 13 '13 at 17:48
  • Another way to see uniqueness is to use the fact that $x \mapsto x^3$ is strictly increasing. – Daniel Fischer Sep 13 '13 at 17:52
  • Aren’t you looking for cube roots of $7+5\sqrt2$? But you probably know already that any two cube roots of a number are related by a factor $(-1\pm\sqrt{-3})/2$. So the other cube roots are not of the desired form. – Lubin Sep 13 '13 at 17:59
  • You may want to look at here. – Samrat Mukhopadhyay Sep 13 '13 at 18:05
  • I find it interesting that this exact problem appears again. I wonder what's the source – Calvin Lin Sep 13 '13 at 18:28

1 Answers1

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The real solutions of a cube root of a positive number are positive numbers, and if you have two positive real-valued solutions then they must be equal. So if $\alpha_1 + \beta_1 \sqrt{2} = \alpha_2 + \beta_2 \sqrt{2}$ are two different solutions where all $\alpha_i, \beta_i$ are rational then you will be able to solve for $\sqrt{2}$ as a rational number, which is impossible. So there is only at most one solution.

user2566092
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