# Find integers $1+\sqrt2+\sqrt3+\sqrt6=\sqrt{a+\sqrt{b+\sqrt{c+\sqrt{d}}}}$

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.

• The first thing I notice is that the left side is $(1+\sqrt 2)(1+ \sqrt 3)$. No guarantee that this helps, but it is probably not an accident. Jun 18 '20 at 4:39
• Shouln't it be $y+z = \sqrt{3 + \sqrt{8}}$ ? Jun 18 '20 at 5:55
• I have a very lengthy method but it's sure to work: keep squaring the lhs until u get to the 4th root Jun 18 '20 at 6:14

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 $$1+\sqrt2+\sqrt3+\sqrt6=\sqrt{21+\sqrt{413+\sqrt{4656+\sqrt{16588800}}}}.$$

First an answer $$1+\sqrt2+\sqrt3+\sqrt6=\sqrt{21+\sqrt{413+\sqrt{4656+ \sqrt{16588800}}}}.$$

Then an explanation.

Everything takes place inside the field $$L=\Bbb{Q}(\sqrt2,\sqrt3)$$. By elementary Galois theory the quadratic subfields of $$L$$ are $$\Bbb{Q}(\sqrt2)$$, $$\Bbb{Q}(\sqrt3)$$ and $$\Bbb{Q}(\sqrt6)$$. The number $$c+\sqrt d$$ must be an element of $$L$$, so we can conclude that $$d=\ell^2 e$$ with some integer $$\ell$$ and $$e\in \{2,3,6\}$$ with the choice of $$e$$ depending on a circumstance we don't know yet.

The key question is the following:

Which elements of $$L$$ have squares in the subfield $$\Bbb{Q}(\sqrt e)$$?

The answer is, again by elementary Galois theory, that for example the square of a number $$z=(a+b\sqrt2+c\sqrt3+d\sqrt6)$$ is in $$\Bbb{Q}(\sqrt6)$$ if and only if either $$a=d=0$$ or $$b=c=0$$. Similarly with the other intermediate fields. This comes from the relevant automorphism of $$L$$ needing to have $$z$$ as an eigenvector belonging to one of the eigenvalues $$+1$$ or $$-1$$.

Let $$\alpha=1+\sqrt2+\sqrt3+\sqrt6$$. Then $$\alpha^2=12+8\sqrt2+6\sqrt3+4\sqrt6.$$ In view of the previous observation we need to find integers $$m,n$$ such that $$(\alpha^2-m)^2-n$$ only contains terms with two of the alternative square roots. Expanding gives $$(\alpha^2-m)^2=m^2-8 \sqrt{6} m-12 \sqrt{3} m-16 \sqrt{2} m-24 m+192 \sqrt{6}+272 \sqrt{3}+336 \sqrt{2}+476.$$ We need one of the square roots to disappear from this by careful choice of $$m$$. Because $$12\nmid 272$$ we cannot make $$\sqrt3$$ disappear. The choice $$m=24$$ would make $$\sqrt6$$ disappear, but then we need to choose $$n=476$$ to kill the coefficient of $$1$$. The catch is that then $$(\alpha^2-24)^2-476<0$$ which is killjoy. It would lead to the answer $$\alpha=\sqrt{24+\sqrt{476-\sqrt{5376+1536 \sqrt{6}}}},$$ but the negative square root is disallowed, I think.

Therefore we must kill the $$\sqrt2$$-terms from $$(\alpha^2-m)^2$$. This forces the choice $$m=21$$, when $$(\alpha^2-21)^2=413+20\sqrt3+24\sqrt6.$$ This, in turn, forces $$n=413$$. As the last step we calculate $$(20\sqrt3+24\sqrt6)^2=4656+2880\sqrt2=4656+\sqrt{16588800}.$$