# Finding a polynomial with integer/rational coefficients and a given algebraic root

Just for fun, I have been attempting to find polynomials with integer coefficients from a given root. I have been able to do this with square roots: for example, if say I wanted to find a polynomial with a root $\sqrt 2$, we would set an equation like this: $$x-\sqrt 2=0$$ Then just move the square root to the other side, square, and end with a result of $$x=\sqrt 2$$ $$x^2=2$$ $$x^2-2=0$$ This strategy can also be done if given a single nth root, or multiple square roots, or a single nth root along with any number of square roots. However, I'm not quite sure how to tackle multiple nth roots like say $3^\frac{1}{3} + 5^\frac{1}{3}$ or $3^\frac{1}{3} + 5^\frac{1}{5}$. Just taking the first example, if we start with the equation $$x-3^\frac{1}{3}-5^\frac{1}{3}=0$$ and move the cube root of 5 to the other side and cube it, we get $$x-3*3^\frac{1}{3}x^2+3*3^\frac{2}{3}x-3=5$$ After getting here I'm unsure of where to go next, as I don't know how to handle both the $3^\frac{1}{3}$ and $3^\frac{2}{3}$ at the same time. I would really appreciate it if someone could help with both this problem and a more general problem solving technique for ones like the other example I showed or something more complicated like say $\frac{3^\frac{2}{3}+5^\frac{1}{5}}{7^\frac{1}{7}+\sqrt 2}$

Let $\,x=\sqrt[3]{3}+\sqrt[3]{5}\,$, then using the identity $\,(a+b)^3=a^3+b^3+3ab(a+b)\,$:

$$x^3 = \left(\sqrt[3]{3}\right)^3+\left(\sqrt[3]{5}\right)^3 + 3 \cdot \sqrt[3]{3}\sqrt[3]{5}\cdot(\sqrt[3]{3}+\sqrt[3]{5}) = 3+5+3 \cdot \sqrt[3]{15} \cdot x = 8 + 3 \sqrt[3]{15} \, x$$

It follows that $\,x^3-8=3 \sqrt[3]{15} \, x\,$, and therefore:

$$(x^3-8)^3=3^3 \cdot 15 \cdot x^3 \quad\iff\quad x^9 - 24 x^6 - 213 x^3 - 512 = 0$$

[ EDIT ] Radicals are algebraic numbers, and the algebraic numbers form a field (i.e. closed under addition/multiplication and their inverses), so any rational expression in algebraic numbers is itself algebraic. A polynomial with integer coefficients having such an expression as a root (though not necessarily the minimal one) can be determined algorithmically using polynomial resultants.

For example, one would write $\,x = \displaystyle\frac{\sqrt[3]{9}+\sqrt[5]{5}}{\sqrt[7]{7}+\sqrt 2} = \frac{a+b}{c+d}\,$ as the polynomial system:

\begin{cases} \begin{align} cx+dx-a-b=0 \\ a^3-9=0 \\ b^5-5=0 \\ c^7-7=0 \\ d^2 - 2 = 0 \end{align} \end{cases}

Then, using resultants repeatedly, the variables $\,a,b,c,d\,$ can be successively eliminated from the system, leaving in the end an equation in $\,x\,$ with integer coefficients. (However, the calculations are not pretty, and would normally be done using some CAS rather than by hand.)

• +1 for answering the question. I suppose I should follow up on my first comment (the one with the inadvertent $abc$ and $xyz$ letter switch in it) so show how the identity I gave can be used with a little more work. – Dave L. Renfro Feb 20 '18 at 20:01
• @DaveL.Renfro Thanks. I thought I had clicked all the posted links, but turns out I somehow missed that very one. Yes, it would make sense to followup, as it's always better to be aware of several lines of proof. – dxiv Feb 20 '18 at 20:07
• Wish the downvoter had left a comment why. – dxiv Feb 23 '18 at 1:06
• Downvote seems strange to me also, since what you did appears correct. You didn't answer the question entirely, but neither did I, at least not in my answer. FYI, I still haven't gotten around to finding a polynomial for my expression . . . maybe I'll get to it later today. (Right now for me is just a quick early a.m. stack exchange check-in before going to the gym.) – Dave L. Renfro Feb 23 '18 at 8:41
• @DaveL.Renfro I thought as well about adding a (simpler but) fully worked out example to my answer using resultants, but then the OP didn't show much interest so I didn't followup on that idea. As for the downvote, I must have rubbed some the wrong way on meta these days, because my answers got more random downvotes all of a sudden than in all of last year. Guess that's just another Friday on MSE ;-) – dxiv Feb 24 '18 at 0:39

What follows is a "school algebra" way to rationalize $\;r_1\sqrt[3]{s_1} + r_2\sqrt[3]{s_2} + r_3\sqrt[3]{s_3},\;$ where $r_1,$ $r_2,$ $r_3,$ $s_1,$ $s_2,$ and $s_3$ are rational numbers. Incidentally, I am not assuming that any of $\sqrt[3]{s_1},$ $\sqrt[3]{s_2},$ or $\sqrt[3]{s_3}$ is irrational.

Note that it suffices to rationalize $\;\sqrt[3]{k} + \sqrt[3]{m} + \sqrt[3]{n},\;$ where $k,$ $m,$ and $n$ are rational numbers (e.g. write $\;r_1\sqrt[3]{s_1} = \sqrt[3]{(r_1)^3s_1},\;$ etc.). In fact, without loss of generality, we can assume $k,$ $m,$ and $n$ are integers, although doing so does not make any difference in what follows. Actually, the reduction to $\sqrt[3]{k} + \sqrt[3]{m} + \sqrt[3]{n}$ is not needed. I only did it to reduce the symbol clutter.

Let $\;a = \sqrt[3]{k}\;$ and $\;b = \sqrt[3]{m}\;$ and $\;c = \sqrt[3]{n}.\;$ Then

$$(a+b+c)(a^2 + b^2 + c^2 - ab - bc - ac) \;\; = \;\; a^3 + b^3 + c^3 - 3abc$$

$$= \;\; (k + m + n) - 3\sqrt[3]{kmn} \;\; = \;\; (k+m+n) -\sqrt[3]{27kmn}$$

Let $\;x = (k+m+n)\;$ and $\;y = \sqrt[3]{27kmn}.\;$ Then

$$(x-y)(x^2 + xy + y^2) \;\; = \;\; x^3 - y^3 \;\; = \;\; (k + m + n)^3 - 27kmn$$

Therefore,

$$\frac{1}{\sqrt[3]{k} + \sqrt[3]{m} + \sqrt[3]{n}}$$

can be rationalized by multiplying both the numerator and the denominator by

$$(a^2 + b^2 + c^2 - ab - bc - ac)(x^2 + xy + y^2)$$

where $a,$ $b,$ $c,$ $x,$ and $y$ are expressions involving $k,$ $m,$ and $n,$ as indicated above. The result will be

$$\frac{(a^2 + b^2 + c^2 - ab - bc - ac)(x^2 + xy + y^2)}{(k + m + n)^3 - 27kmn}$$

• I am not assuming that any of ... is irrational (+1) Obvious as it may be, I would perhaps emphasize that this makes it directly applicable to the given $\,x-\sqrt[3]{3}-\sqrt[3]{5}\,$. – dxiv Feb 20 '18 at 21:40
• @dxiv: Looking at this just now, the next (very early) morning, I realize that I didn't actually answer the question, which is how to find a polynomial with integer coefficients having the given number as a root. It's only requires some minor adjustments to what I gave, but I don't have time to deal with it now. I'll try to get to it at some later time. When I get to it, I'll keep what I have, and include an amendment that gives the desired polynomial. – Dave L. Renfro Feb 21 '18 at 7:54
• @dxiv: I'm not sure anymore that what I did can be used to easily obtain a Z-polynomial having $\sqrt[3]{k}+\sqrt[3]{m}+\sqrt[3]{n}$ as a root. I have a huge collection of papers (many from the late 1700s through the 1800s) on rationalizing methods, but I'm too busy with other things to get more involved in this now. Of possibly interest is this 4 December 1999 sci.math post of mine (Salvatore Composto's papers I cite can now be found on the internet by googling, I think), and Stan Brown's notes. – Dave L. Renfro Feb 24 '18 at 14:29