# Proof that N = (45+29 sqrt(2))^(1/3)+(45-29 sqrt(2))^(1/3) is a integer ?

Sep 13, 2016

Consider ${t}^{3} - 21 t - 90 = 0$

This has one Real root which is $6$ a.k.a. ${\left(45 + 29 \sqrt{2}\right)}^{\frac{1}{3}} + {\left(45 - 29 \sqrt{2}\right)}^{\frac{1}{3}}$

#### Explanation:

Consider the equation:

${t}^{3} - 21 t - 90 = 0$

Using Cardano's method to solve it, let $t = u + v$

Then:

${u}^{3} + {v}^{3} + 3 \left(u v - 7\right) \left(u + v\right) - 90 = 0$

To eliminate the term in $\left(u + v\right)$, add the constraint $u v = 7$

Then:

${u}^{3} + {7}^{3} / {u}^{3} - 90 = 0$

Multiply through by ${u}^{3}$ and rearrange to get the quadratic in ${u}^{3}$:

${\left({u}^{3}\right)}^{2} - 90 \left({u}^{3}\right) + 343 = 0$

by the quadratic formula, this has roots:

${u}^{3} = \frac{90 \pm \sqrt{{90}^{2} - \left(4 \cdot 343\right)}}{2}$

$\textcolor{w h i t e}{{u}^{3}} = 45 \pm \frac{1}{2} \sqrt{8100 - 1372}$

$\textcolor{w h i t e}{{u}^{3}} = 45 \pm \frac{1}{2} \sqrt{6728}$

$\textcolor{w h i t e}{{u}^{3}} = 45 \pm 29 \sqrt{2}$

Since this is Real and the derivation was symmetric in $u$ and $v$, we can use one of these roots for ${u}^{3}$ and the other for ${v}^{3}$ to deduce that the Real zero of ${t}^{3} - 21 t - 90$ is:

${t}_{1} = \sqrt{45 + 29 \sqrt{2}} + \sqrt{45 - 29 \sqrt{2}}$

but we find:

${\left(6\right)}^{3} - 21 \left(6\right) - 90 = 216 - 126 - 90 = 0$

So the Real zero of ${t}^{3} - 21 t - 90$ is $6$

So $6 = \sqrt{45 + 29 \sqrt{2}} + \sqrt{45 - 29 \sqrt{2}}$

$\textcolor{w h i t e}{}$
Footnote

To find the cubic equation, I used Cardano's method backwards.

Sep 13, 2016

$N = 6$

#### Explanation:

Making $x = 45 + 29 \sqrt{2}$ and $y = 45 - 29 \sqrt{2}$ then

${\left({x}^{\frac{1}{3}} + {y}^{\frac{1}{3}}\right)}^{3} = x + 3 {\left(x y\right)}^{\frac{1}{3}} {x}^{\frac{1}{3}} + 3 {\left(x y\right)}^{\frac{1}{3}} {y}^{\frac{1}{3}} + y$

${\left(x y\right)}^{\frac{1}{3}} = {\left({7}^{3}\right)}^{\frac{1}{3}} = 7$
$x + y = 2 \times 45$

so

${\left({x}^{\frac{1}{3}} + {y}^{\frac{1}{3}}\right)}^{3} = 90 + 21 \left({x}^{\frac{1}{3}} + {y}^{\frac{1}{3}}\right)$

or calling $z = {x}^{\frac{1}{3}} + {y}^{\frac{1}{3}}$ we have

${z}^{3} - 21 z - 90 = 0$

with $90 = 2 \times {3}^{2} \times 5$ and $z = 6$ is a root so

${x}^{\frac{1}{3}} + {y}^{\frac{1}{3}} = 6$