Algebraic Numbers

As in the example given of $\sqrt{2}$ , it is apparent that $\sqrt{2+\sqrt{3}+\sqrt{5}}$ is a root of the irrational polynomial $x^2-(2+\sqrt{3}+\sqrt{5})$

For any $x$ that is a solution to $x^2-2-\sqrt{3}-\sqrt{5}=0$ , we necessarily have that $(x^2-2-\sqrt{3})^2-5=0$ .

Expanding, simplifying and isoating the $\sqrt{3}$ term, we get $\frac {x^4-4x^2+2} {2x^2-4} -\sqrt{3}=0$ at which point we repeat the step earlier and see that $(\frac {x^4-4x^2+2} {2x^2-4})^2-3=0$

Expanding and simplifying leads us to $x^8-8x^6+8x^4+32x^2-44=0$ and from here simple addition gives that $0+|-8|+0+|8|+0+|32|+0+|-44|=92$

$x-\sqrt{2+\sqrt{3+\sqrt{5}}}$ (1) To remove the radicals in (1), multiply it by $x+ \sqrt{2+\sqrt{3+\sqrt{5}}}$ . The resulting expression is $x^2-(2+\sqrt{3+\sqrt{5}})$ (2) Then multiply (2) by $x^2+(-2+\sqrt{3+\sqrt{5}})$ to remove more radicals. The resulting expression is $x^4-4x^2-(4+2\sqrt{15})$ (3) And using the same technique in removing the radicals in (2), multiply (3) by $(x^4-4x^2)+(-4+2√15)$ . The resulting expression is $x^8-8x^6+8x^4+32x^2-44$ . Then, adding the absolute values of the coefficients of the terms (not including the term $x^8$ ), the result is 92, the desired answer.

[Latex edits. Note that $\sqrt{3} + \sqrt{5} \neq \sqrt{3 + \sqrt{5}}$ . Be clear in your writeup - Calvin]

Let $x = \sqrt{ 2 + \sqrt{3} + \sqrt{5} }$ . Then $x^2 - 2 = \sqrt{3} + \sqrt{5}$ , so $(x^2 - 2)^2 = 8 + 2 \sqrt{15}$ and so $((x^2-2)^2-8)^2 - 60 = 0$ . The polynomial we seek is $x^8 - 8x^6 + 8x^4 + 32x^2 - 44$ , so the answer is 92.

This repeated squaring is the only way to get rid of the square roots, and it takes three squarings to get rid of all square roots, so the polynomial which has $\sqrt{ 2 + \sqrt{3} + \sqrt{5} }$ as a root must have degree 8. This monic polynomial must be unique, for if $p(x)$ and $q(x)$ were different monic polynomials that had $\sqrt{ 2 + \sqrt{3} + \sqrt{5} }$ as a root, then $p(x) - q(x)$ would be a polynomial of degree less than 8 which would still have $\sqrt{ 2 + \sqrt{3} + \sqrt{5} }$ as a root.

In general, a number is still called algebraic if there is a polynomial with rational coefficients for which the number is a root. The minimal polynomial of an algebraic number is the unique monic polynomial of smallest degree (with rational coefficients) which has the number as a root. Since the minimal polynomial of $\sqrt{ 2 + \sqrt{3} + \sqrt{5} }$ actually has integer coefficients, the number $\sqrt{ 2 + \sqrt{3} + \sqrt{5} }$ is called an algebraic integer .