TITLE: Find integers $1+\sqrt2+\sqrt3+\sqrt6=\sqrt{a+\sqrt{b+\sqrt{c+\sqrt{d}}}}$
QUESTION [7 upvotes]: Root numbers Problem (Math Quiz Facebook):

Consider the following equation:
$$1+\sqrt2+\sqrt3+\sqrt6=\sqrt{a+\sqrt{b+\sqrt{c+\sqrt{d}}}}$$
Where $a,\,b,\,c,\,d$ are integers. Find $a+b+c+d$

I've tried it like this:
Let $w=\sqrt6,\, x=\sqrt3, \, y=\sqrt2, z=1$
$$\begin{align}
(y+z)^2 &= (y^2 + z^2) + 2yz\\
y+z &= \sqrt{(y^2 + z^2) + 2yz}\\
y+z &=  \sqrt{3 + \sqrt{8}}
\end{align}$$
Let $y+z=f$
$$\begin{align}
(x+f)^2 &= (x^2 + f^2) + 2xf\\
x+f &= \sqrt{(x^2 + f^2) + 2xf}\\
x+f &=  \sqrt{(9+\sqrt8) + 2\sqrt{9+3\sqrt8}}
\end{align}$$
And I don't think this going to work since there's still a root term on the bracket that is $9+\sqrt8$. I need another way to make it as an integer.

REPLY [3 votes]: Expand out enough to get to
\begin{align*}
(a^2-24a+476-b)+\sqrt{2}(336-16a)+\sqrt{3}(272-12a)+\sqrt{6}(192-8a)&=\sqrt{c+\sqrt{d}}.
\end{align*}
This means, when we square the left side, we need to only have two terms with nonzero coefficient. Note that
$$(w+x\sqrt2+y\sqrt3+z\sqrt6)^2=(w^2+2x^2+3y^2+6z^2)+2\sqrt2(wx+3yz)+2\sqrt3(wy+2xz)+2\sqrt6(wz+xy),$$
so we need two of $\{wx+3yz,wy+2xz,wz+xy\}$ to be $0$. However, if the first two are $0$, then
$$wxy+3y^2z=wxy+2x^2z=0$$
implies that either $z=0$ or $x=y=0$; in the first case, $w=0$. We may get similar conclusions for each of the other selections to be $0$, so we must have that two of the parameters $\{w,x,y,z\}$ are $0$. In particular, since none of our polynomials in $a$ for $x,y,z$ have common roots, we must have that $w=0$. Then, $y\neq 0$ since $y$ has a noninteger root for $a$, so we have $a\in\{21,24\}$ and $a=21\implies b=413$, with $a=24\implies b=476$. If $a=24$, the left side is actually negative (it's $-48\sqrt2-16\sqrt3$), so it can't be the square root of anything. For $a=21$, $b=413$, we may find by direct calculation that
$$1+\sqrt2+\sqrt3+\sqrt6=\sqrt{21+\sqrt{413+\sqrt{4656+\sqrt{16588800}}}}.$$
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