Minimal Polynomial of $\sqrt{2}+\sqrt{3}+\sqrt{5}$ To find the above minimal polynomial, let 
$$x=\sqrt{2}+\sqrt{3}+\sqrt{5}$$
$$x^2=10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}$$
Subtracting 10 and squaring gives
$$x^4-20x^2+100=4(31+2\sqrt{60}+2\sqrt{90}+2\sqrt{150})$$
$$x^4-20x^2+100=4(31+4\sqrt{15}+6\sqrt{10}+10\sqrt{6})$$
$$x^4-20x^2-24=40\sqrt{6}+24\sqrt{10}+16\sqrt{15}$$
$$x^4-20x^2-24=8(2\sqrt{6}+2\sqrt{10}+2\sqrt{15})+24\sqrt{6}+8\sqrt{10}$$
$$x^4-20x^2-24=8(x^2-10)+24\sqrt{6}+8\sqrt{10}$$
$$x^4-28x^2-104=24\sqrt{6}+8\sqrt{10}$$
Again, squaring both sides
$$x^8-56x^6+576x^4+5428x^2+10816=4096+765\sqrt{6}$$
But if I square again, I will get a degree 16 polynomial.  Mathematica says the minimal polynomial is degree 8, which would make sense since elements of $\mathbb{Q}[\sqrt{2},\sqrt{3},\sqrt{5}]$ look like
$$a+b\sqrt{2}+c\sqrt{3}+d\sqrt{5}+e\sqrt{6}+f\sqrt{10}+g\sqrt{15}+h\sqrt{30}$$
Where am I making mistakes?
 A: The 'easy' way to get at the minimal polynomial in this case is to take the product of all the terms $x\pm\sqrt2\pm\sqrt3\pm\sqrt5$; this is, in essence, because the Galois group over $\mathbb{Q}$ in this case is just $(\mathbb{Z}/2\mathbb{Z})^3$, with each of the three copies of $\mathbb{Z}/2\mathbb{Z}$ corresponding to a sign change on one of the three square root terms.  The simple approach (take the product of $(x-\alpha)$ over all of the possible $\alpha$ obtained by applying the automorphisms in the Galois group to $\sqrt2+\sqrt3+\sqrt5$) works here because $\sqrt2$, $\sqrt3$ and $\sqrt5$ are 'mutually irreducible' over $\mathbb{Q}$ (in the sense that $\sqrt2\not\in\mathbb{Q}(\sqrt{3}, \sqrt{5})$, etc.) , so we can compose extensions in a clean fashion.
A: I started out by observing that $2+3 = 5$; I thought that might make it easier.  Then
$$
x-\sqrt{5} = \sqrt{2}+\sqrt{3}
$$
$$
x^2-2\sqrt{5}x+5 = 2\sqrt{6}+5
$$
$$
x^2-2\sqrt{5}x = 2\sqrt{6}
$$
$$
x^4-4\sqrt{5}x^3+20x^2 = 24
$$
$$
x^4+20x^2-24 = 4\sqrt{5}x^3
$$
$$
x^8+40x^6+352x^4-960x^2+576 = 80x^6
$$
$$
x^8-40x^6+352x^4-960x^2+576 = 0
$$
(Or, of course, you can do it more systematically via the Galois theory!)
A: I thought I'd get rid of the surds one-by-one: first get rid of $\sqrt2$, then $\sqrt3$, and finally $\sqrt5$. To get rid of a surd, I would get everything with that surd in it to one side, and then square. (Note that $\sqrt{15}$ has a $\sqrt3$ in it.)
\begin{align}
x&=\sqrt2+\sqrt3+\sqrt5\\
x-\sqrt3-\sqrt5&=\sqrt2\\
(x-\sqrt3-\sqrt5)^2&=2\\
x^2+(-2\sqrt5-2\sqrt3)x+(2\sqrt{15}+8)&=2\\
x^2+(-2\sqrt5-2\sqrt3)x+(2\sqrt{15}+6)&=0\\
x^2-2\sqrt5x+6&=2\sqrt3x-2\sqrt{15}\\
(x^2-2\sqrt5x+6)^2&=(2\sqrt3x-2\sqrt{15})^2\\
x^4-4\sqrt5x^3+32x^2-24\sqrt5x+36&=12x^2-24\sqrt5x+60\\
x^4-4\sqrt5x^3+20x^2-24&=0\\
x^4+20x^2-24&=4\sqrt5x^3\\
(x^4+20x^2-24)^2&=80x^6\\
x^8+40x^6+352x^4-960x^2+576&=80x^6\\
x^8-40x^6+352x^4-960x^2+576&=0
\end{align}
This, of course, is equal to the Galois theory answer of $\displaystyle\prod(x\pm\sqrt2\pm\sqrt3\pm\sqrt5)$, where you multiply every possible choice of signs together.
A: An elementary way 
$(x-\sqrt2)^2=(\sqrt3+\sqrt5)^2 \implies x^2+2-2x\sqrt2=8+2\sqrt{15}\implies(x^2-6-2x\sqrt2)^2=((x^2-6)^2-4x(x^2-6)\sqrt2+8x^2)^2=60\implies((x^2-6)^2+8x^2-60)^2=(4x(x^2-6)\sqrt2)^2$ 
This resulting polynomial of $\mathbb{Q}[x]$ is obviously of degree 8
A: In general, if $p_1,\dotsc,p_n$ is any list of integers, then the polynomial
$$f=\prod_{e_1,\dotsc,e_n \in \{\pm 1\}} (x+e_1 \sqrt{p_1}+\dotsc+e_n \sqrt{p_n})$$
has coefficients in $\mathbb{Z}$, which follows by induction from the observation

$p \in \mathbb{Z},\,g \in \mathbb{Z}[x] \Rightarrow g(\sqrt{p}) \cdot g(-\sqrt{p}) \in \mathbb{Z}[x]$.

You can prove this using the automorphism group of $\mathbb{Z}[\sqrt{p}]$. But you can also just verify it via some direct calculation, which has the advantage that you really see why the $\sqrt{p}$-terms vanish and that you may compute $f$ faster.
Clearly, $f$ is monic, has degree $2^n$, and has $\sqrt{p_1}+\dotsc+\sqrt{p_n}$ as a root. So this is quite elementary and for any fixed $n$, you can also compute $f$. What is not so easy to prove is that if $p_1,\dotsc,p_n$ are square-free and pairwise coprime integers, then $f$ is irreducible. Equivalently, the degree of $\sqrt{p_1}+\dotsc+\sqrt{p_n}$ equals $2^n$. You can find the proof here ("square-root-extension.pdf").
For the numbers $p_1,p_2,p_3 = 2,3,5$, we get $f = x^8 - 40 x^6 + 352 x^4 - 960 x^2 + 576$.
