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Let $$x=\sqrt{a + \sqrt{b}} + \sqrt{c + \sqrt{d}}$$

How do you find the polynomial with this value as a root? Where a, b, c, and d are integers.

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    Square things, carefully, several times.2012-10-07
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    If my answer is not satisfying enough I can give you an idea of how to construct the polynomial who has the sum as a root. I just thought you might've wanted a hint first.2012-10-07
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    I wrote [a blog article about how to do this](http://blog.plover.com/math/degree.html) that you might find helpful. It includes an example of how to do it for $\sqrt2+\sqrt3$. This example is simpler than yours, but the process is similar.2012-10-07
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    @MJD, +1 for interesting article, thanks for sharing.2012-10-07
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    Does the squaring really work. It seems that the cross product term gets uglier and uglier. Do you see it simplifying eventually.2012-10-07
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    Yes, I think it does, but you need to be careful about what you're squaring.2012-10-07
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    @Qiaochu : I don't think that simply nesting " "$^2$ will eventually simplify things... I thought that "carefully squaring" meant that you subtracted stuff along the way to make sure things work. Did you literally mean "Square repeatedly and keep track of the terms"? I don't think it would work.2012-10-07
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    @Patrick: where did I claim that nesting squares would work?2012-10-07
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    @Qiaochu : You didn't, but I didn't know how to interpret your comments, they seemed quite vague (to purposely let your comment be hints, of course). Just don't mind ; we probably agree on what to do anyway.2012-10-07
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    @Patrick: I have given (most of) the solution I was hinting at below.2012-10-07
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    @Qiaochu Yuan : Yeah okay my approach was different. I was finishing my argument using the root of polynomial <-> eigenvalue of matrix to finish the argument and then use the argument to find a root of $\alpha + \beta$ using the matrix of $\alpha$ and $\beta$ plus properties of tensor products of matrices. Probably a not-very-elementary construction though, although I don't think the computations are very hard.2012-10-07

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A hint would be to first find the polynomial for $\sqrt{a + \sqrt b}$ first (thus also giving you a formula for the second term), and then find the polynomial with $\alpha + \beta$ as a root given polynomials $f$ and $g$ such that $f(\alpha) = g(\beta) = 0$. I think that will make your search a little bit easier.

How about this one for the first term? $$ (x^2 - a)^2 - b $$

EDIT : Since you asked for the proof, I will outline it here.

Note that for a polynomial $p(x) = a_0 + \dots + a_{n-1}x^{n-1} + x^n$, being a root of this polynomial is equivalent to being an eigenvalue of the following matrix called the companion matrix of $p$ : $$ \begin{bmatrix} 0 & & & & - a_0 \\ 1 & 0 & & & \vdots \\ & 1 & \ddots & & \vdots \\ & & \ddots & 0 & -a_{n-2} \\ & & & 1 & -a_{n-1} \\ \end{bmatrix} $$ You can read it off Dummit & Foote's Abstract Algebra, chapter 12. It is a very beautiful theory I suggest you know if you study polynomials for a while.

So instead of looking for a polynomial with $\alpha + \beta$ as a root, we are looking for a matrix with $\alpha + \beta$ as its eigenvalue, because then the characteristic polynomial of that matrix will give us the desired polynomial.

To produce such a matrix, one will use the tensor product of matrices (Very pretty schemes there explain how to perform such an operation : http://en.wikipedia.org/wiki/Tensor_product#Kronecker_product_of_two_matrices) because they have very interesting properties. Say $\alpha$ is an eigenvalue of the $n \times n$ matrix $A$ with eigenvector $x$ and $\beta$ is an eigenvalue of the $m \times m$ matrix $B$ with eigenvector $y$. Denote by $I_n$ the $n\times n$ identity matrix. Then $$ (A \otimes I_m + I_n \otimes B)(x \otimes y) = (Ax) \otimes y + x \otimes (By) = (\alpha x) \otimes y + x \otimes (\beta y) = (\alpha)(x \otimes y) + (\beta) (x \otimes y) = (\alpha + \beta) (x \otimes y) $$ so that $\alpha + \beta$ is an eigenvalue of the matrix $A \otimes I_m + I_n \otimes B$.

If we do the computations explicitly, the matrix associated to $(x^2 - a)^2 - b$ will be a $4 \times 4$ matrix, namely since the polynomial is $x^4 - 2ax^2 + a^2 - b$, the matrix will be $$ \begin{bmatrix} 0 & 0 & 0 & -a^2 + b \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 2a \\ 0 & 0 & 1 & 0 \\ \end{bmatrix} $$ and similarly the matrix for $B$ will be $$ \begin{bmatrix} 0 & 0 & 0 & -c^2 + d \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 2c \\ 0 & 0 & 1 & 0 \\ \end{bmatrix}. $$ Now computing the tensor products gives $$ A \otimes I_4 + I_4 \otimes B = \begin{bmatrix} 0 & 0 & 0 & -c^2 + d & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -a^2 + b & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -a^2 + b & 0 & 0 \\ 0 & 1 & 0 & 2c & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -a^2 + b & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -a^2 + b \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & -c^2 + d & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 & 0 & 2c & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -c^2 + d & 2a & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 2a & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 2c & 0 & 0 & 2a & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 2a \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -c^2 + d \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 2c \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \\ \end{bmatrix} $$ So compute the characteristic polynomial of this guy and you're good to go. If you're interested, you could try to plug it into Mathematica and see if trivial factors come out. Degree $16$ feels too high for the degree of this guy, but maybe it just is, you know. =)

Hope that helps,

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    Let me ponder this hint for a day or so. Thanks!2012-10-07
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    It'll help if you know how to find a polynomial with $\alpha + \beta$ as a root when you know polynomials for $\alpha$ and $\beta$. Have you ever heard of such a technique?2012-10-07
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    No, but it will be fun to think about.2012-10-07
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    @NoOneinParticular : It will be long, too. The only proof I know uses tensor products of matrices and it is a very clever argument (simple to read though).2012-10-07
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    In that case, do you have a pointer to the paper?2012-10-08
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    There's only two lines missing, my matrix was wayyyy too large. But it is a $16 \times 16$ matrix after all... the first two columns only contain $1$'s and $0$'s, click on edit if you wanna take a look at them.2012-10-08
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    Outstanding answer. I really appreciate it.2012-10-08
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    @NoOneinParticular : Just for fun, did you compute the big guy?2012-10-09
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    Still trying to get it right. (Doing it by hand.) I will post the result when I get it done.2012-10-14
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For the sake of defending my honor, here's most of the solution I alluded to involving squaring. It is somewhat messy, and the solution alluded to by Patrick Da Silva is nicer.

First square:

$$x^2 = a + \sqrt{b} + 2 \sqrt{(a + \sqrt{b})(c + \sqrt{d})} + c + \sqrt{d}.$$

Second square:

$$(x^2 - a - \sqrt{b} - c - \sqrt{d})^2 = 4 (a + \sqrt{b})(c + \sqrt{d}).$$

Expand:

$$x^4 - 2x^2 (a + \sqrt{b} + c + \sqrt{d}) + (a^2 + 2a \sqrt{b} + b) + 2(ac + c \sqrt{b} + a \sqrt{d} + \sqrt{bd}) + c^2 + 2c \sqrt{d} + d = 4(ac + c \sqrt{b} + a \sqrt{d} + \sqrt{bd})$$

Rearrange:

$$x^4 - 2x^2 (a + c) + (-2x^2 + 2a - 2c) \sqrt{b} + (a^2 + b - 2ac + c^2 + d) = (2x + 2a - 2c) \sqrt{d} + 2 \sqrt{bd}.$$

I won't write everything out from here (I've probably already made a mistake). Squaring a third time removes all of the radicals except those of the form $\sqrt{b}$. Rearranging and squaring a fourth time removes these.

This method does not generalize; putting too many additional square roots into the problem will cause squaring to keep giving you more terms regardless of how cleverly you rearrange.

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    Thanks. I was looking at keeping all of the radicals on the right. How did you decide which to move to the left.2012-10-07
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    I'm flattered =)2012-10-07
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    @No One: experience. The first step is to get rid of the nested radicals by any means necessary. The second step is motivated by some knowledge of the theory of field extensions (the keyword here is "biquadratic extension").2012-10-07
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    @Qiaochu Yuan: "This method does not generalize; putting too many additional square roots ..." I think things can be organized in such a way so that, no matter how many radicals you start with, you can eventually remove all of them. The key is to observe that starting with $n$ many radicals, there is a fixed upper bound on the number "irrational terms" that can show up. For example, if $a,$, $b,$ $c,$ $d,$ and $e$ are each originally under a radical, then under radicals you'll only get these, and things like $ab,$ $ac,$ ..., and things like $abc,$ $abd,$ ... $abcd,$ etc. under radicals.2012-10-08
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    @Qiaochu Yuan: More specifically, at most $2^{n}- 1$ many different square root terms can show up, this being the number of different singleton to $n$-fold unordered products of the insides of the radicals, and this number of such distinct products is equal to $C(n,1) + C(n,2) + ... + C(n,n),$ where $C(n,k)$ is the binomial coefficient for the number of $k$-element subsets of a set with $n$ elements.2012-10-08
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    @Dave: it's true that there are only boundedly many terms that can occur, but I don't think you can remove them by just squaring. I admit that I have not tried to do this myself, but see Iurie Boreico's comments on this method at http://www.thehcmr.org/issue2_1/mfp.pdf .2012-10-08
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    @Qiaochu Yuan: Unfortunately, I have to leave in a few minutes, but I still think we can do this. Systematically clear out each of the terms such as $\sqrt{a}$ and $\sqrt{b},$ by putting all terms and only those terms involving one of them on one side and squaring. Then start working on terms such as $\sqrt{ab}.$ You can't pick up any terms whose irrational factor is $\sqrt{a}$ by squaring other stuff, so once you get rid of them, they're gone. But I think you have to do the single irrational factor first, then the double irrational factors, etc.2012-10-08
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    @Dave: unfortunately, you can absolutely pick up a term whose irrational factor is $\sqrt{a}$ by squaring other stuff, such as $\sqrt{b}$ and $\sqrt{ab}$.2012-10-08
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    @Qiaochu Yuan: I think the method I'm describing is outlined in my reference [6] (Daniel Mooney's 1797 paper in **Transactions of the Royal Irish Academy**) at [my 16 November 2010 AP-Calculus post](http://mathforum.org/kb/message.jspa?messageID=7288329). Note that my post describes another way, one that involves multiplying by a certain polynomial, and I realize that's not the "square and conquer" method we're talking about.2012-10-08
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    @Qiaochu Yuan: I'm sorry, but I'm late as it is. I'll have to look at this tomorrow. For now I'll say that there might be a distinction between working purely algebraically and working with radicals of integers, which can be factored into square-free factors (or something like this).2012-10-08
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    @Qiaochu Yuan: Suppose, for example, the irrational terms are $\sqrt{a},$ $\sqrt{b},$ $\sqrt{c},$ $\sqrt{d},$ and $\sqrt{e}.$ Note this is more than Iurie Boreico says can be removed in his manuscript that you gave a link to. Put $\sqrt{a}$ on the left side and everything else on the right side and square. Now put all terms of the form (stuff not involving $b$)$\cdot \sqrt{b}$ on the left side and everything else on the right side and square. Now do the same thing for $c$ that we did for $b.$ Keep going, and after at most $5$ squarings all radicals will be removed.2012-10-09