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.
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.
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,
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.