Converting sums of square-roots to nested square-roots

All difference between numbers of the form $$x=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{5}\qquad\qquad (a,b,c,d\in\mathbb{Z}_+)$$ and $$y=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}\qquad\qquad (a,b,c,d\in\mathbb{Z}_+)$$ is that their powers can be written (uniquely) as $$ x^p = k_{p1} + k_{p2}\sqrt{2} + k_{p3}\sqrt{3} + k_{p4}\sqrt{5} + k_{p5}\sqrt{6} + k_{p6}\sqrt{10} + k_{p7}\sqrt{15} +k_{p8}\sqrt{30}, \tag{1} $$ but $$ y^p = l_{p1} + l_{p2}\sqrt{2} + l_{p3}\sqrt{3} + l_{p4}\sqrt{6}, \tag{1'} $$ where $k_{pj}\in\mathbb{Z}_+$, $l_{pj}\in\mathbb{Z}_+$.

---

If one want to write $x$ in the form of nested radicals $$ x=\sqrt{a_1+\sqrt{a_2+\sqrt{...+\sqrt{a_{n-1}+\sqrt{a_n}}}}}, $$

then $x$ is the root of the polynomial of $2^n$-th degree $$ P(x) = \left(\left(\left(x^2-a_1\right)^2-a_2\right)^2-\cdots-a_{n-1}\right)^2-a_n, $$ $$ P(x) = x^{2^n} + p_{2^{n-1}-1}(a_1,...,a_n)x^{2^n-2}+\cdots + p_1 (a_1,...,a_n)x^2 + p_0(a_1,...,a_n), $$ $$ P(x) = \sum_{j=0}^{2^{n-1}} p_{j}(a_1,...,a_n)x^{2j}, $$ where $p_{j}(a_1,...,a_n)$ are polynomials of $a_1,...,a_n$ with integer coefficients.

Let's split polynomial $P(x)$ into $8$ linear (rational) independent parts:

$$ P(x) =\\ 1\cdot \sum_{j=0}^{2^{n-1}}p_j(a_1,...,a_n)k_{2j,1} \\ + \sqrt{2} \cdot \sum_{j=0}^{2^{n-1}}p_j(a_1,...,a_n)k_{2j,2} \\ + \sqrt{3} \cdot \sum_{j=0}^{2^{n-1}}p_j(a_1,...,a_n)k_{2j,3} \\ + \sqrt{5} \cdot \sum_{j=0}^{2^{n-1}}p_j(a_1,...,a_n)k_{2j,4} \\ + \cdots \\ + \sqrt{30} \cdot \sum_{j=0}^{2^{n-1}}p_j(a_1,...,a_n)k_{2j,8}.\tag{2} $$

Sum in each row must be equal to $0$.

But each sum as polynomial of $a_1,...,a_n$ depends on $n$ integer variables/values.

$8$ (non-linear) equations for $n$ values.

There must be very lucky coincidence if $4$ variables $a_1,a_2,a_3,a_4$ will support all $8$ equations.

---

I think, there must be $8$ nested radicals here to provide as many "free" variables $a_1,...,a_n$:

$$ a+b\sqrt{2}+c\sqrt{3}+d\sqrt{5} = \sqrt{a_1+\sqrt{a_2+\sqrt{...+\sqrt{a_7+\sqrt{a_8}}}}}. $$

It is hard to find such form at all.

---

Note: if $x$ is of the form $$ x = b\sqrt{2}+c\sqrt{3}+d\sqrt{5}, $$ then (see $(1)$) $$ x^{2j} = k_{2j,1}+k_{2j,5}\sqrt{6}+k_{2j,6}\sqrt{10}+k_{2j,7}\sqrt{15}, $$ (coefficients near $\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{30}$ are $0$), so there are only $4$ rows in $(2)$.

A few examples of radicals of such form:

$$ \sqrt{2}+\sqrt{3}+\sqrt{7} = \sqrt{16+\sqrt{180+\sqrt{10\;496 + \sqrt{34\;406\;400}}}}, $$

$$ \sqrt{2}+2\sqrt{3}+\sqrt{5} = \sqrt{23+\sqrt{392+\sqrt{77\;824+\sqrt{3\;538\;944\;000}}}}, $$

$$ 3\sqrt{2}+\sqrt{3}+\sqrt{5} = \sqrt{32+\sqrt{672+\sqrt{229\;824+\sqrt{11\;943\;936\;000}}}}. $$

Writing a number $\alpha$ in this nested square root form $$\sqrt{a_1+\sqrt{a_{2}+\sqrt{...\sqrt{a_n}}}} \; \; (1)$$ implies that $\alpha$ satisfies the equation $$(((x^2-a_1)^2-a_2)^2...)^2-a_n = 0 \; \; (2)$$ where the $a_i$ are integers. In fact, allowing both values for the square root, the numbers of the form (1) are precisely those satisfying equation (2).

Expanding out (2), we find that it is a monic polynomial in $x^2$ of degree $2^{n-1}$ with integer coefficients. For $n \ge 3$, the polynomial will have more than $n$ coefficients after the first, so these cannot be freely specified via choosing $a_1,...,a_n$. We do get, however, that if number $\alpha$ can be written in the form (1) then $\alpha^2$ is a root of a monic integer polynomial of degree $2^k$ for some $k \in \{0,1,2,...\}$.

If you generalise the form to $$a_1+\sqrt{a_2+\sqrt{a_{3}+\sqrt{...\sqrt{a_n}}}} \; \; (1*)$$ then the necessary condition becomes that $\alpha$ itself is a root of such a polynomial.

Roots of monic integer polynomials (of any degree) are called algebraic integers. See http://en.wikipedia.org/wiki/Algebraic_integer

For any pair $(a,b)$ one can write $$ \sqrt{a}+\sqrt{b} = \sqrt{A+2\sqrt{B}},\tag{1} $$ where
$A=a+b$,
$B=ab$.

---

For any ordered triple $(a,b,c)$, such that $1\le a\le b \le c$, one can write $$ \sqrt{a}+\sqrt{b}+\sqrt{c} = \sqrt{A +2\sqrt{B+2\left(\sqrt{p}+\sqrt{q}\right)}} = \sqrt{A + 2\sqrt{B + 2\sqrt{C + 2\sqrt{D}}}},\tag{2} $$

where
$A=3a+b+c$,
$B=(a+b)(a+c)$,
$p = ac(b-a)^2$,
$q = ab(c-a)^2$,
$C = p+q$, (see $(1)$)
$D = pq$.

Note, that
$\sqrt{p} = (b-a)\sqrt{ac}$,
$\sqrt{q} = (c-a)\sqrt{ab}$,
$\sqrt{D} = a (b-a) (c-a) \sqrt{bc}$.

---

Maybe there is reason to search $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}$ in the form $$\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}= \sqrt{A+2\sqrt{B+2\sqrt{C+2\sqrt{D+2\left(\sqrt{p}+\sqrt{q}+\sqrt{r}\right)}}}},\tag{3}$$ and (if success) apply $(2)$ to the sum $\sqrt{p}+\sqrt{q}+\sqrt{r}$.

Here is other approach. I hope it is most helpful.

A.
For any vector $v$ with integer coefficients $$ v = (a,b,c,d,e,f,g,h) $$ denote linear combination $$ x_v = a+b\sqrt{2}+c\sqrt{3}+d\sqrt{5}+e\sqrt{6}+f\sqrt{10}+g\sqrt{15}+h\sqrt{30}.\tag{1} $$

Values $x_v$ generate ring (all sums, products of them has form $(1)$ etc).

Value $$ x_v^2-A $$ has according/appropriate vector with integer coefficients (vector $w=(a',b',c',d',e',f',g',h')$, where its coefficients are defined as $$ \begin{array}{l} a'=a^2+2b^2+3c^2+5d^2+6e^2+10f^2+15g^2+30h^2-A;\\ b'=2(ab+3ce+5df+15gh),\\ c'=2(ac+2be+5dg+10fh),\\ d'=2(ad+2bf+3cg+6eh),\\ e'=2(ae+bc+5fg+5dh),\\ f'=2(af+bd+3eg+3ch),\\ g'=2(ag+cd+2ef+2bh),\\ h'=2(ah+bg+cf+de).\tag{2} \end{array} $$

So, one can define vector transformation $$ w = T(v,A),\tag{3} $$ described by $(2)$.

---

B.
If there is representation $$ 3+\sqrt{2}+\sqrt{3}+\sqrt{5}=\sqrt{A_1+\sqrt{A_2+\sqrt{A_3+...+\sqrt{A_{n-1}+\sqrt{A_n}}}}},\tag{4} $$ then for vector $v_0=(3,1,1,1,0,0,0,0)$ exist these transformations:

$$ v_1=T(v_0,A_1),\\ v_2=T(v_1,A_2),\\ ...\\ v_n=T(v_{n-1},A_n),\tag{5} $$ where $v_n$ iz zero-vector: $v_n=(0,0,0,0,0,0,0,0)$.

---

C.

Now, apply modular arithmetic to $(2), (3), (5)$.

For $m\in\mathbb{N}$ denote transformation

$$ z = T_m(v,A),\tag{3'} $$ where $z=(a'\bmod m, ~ b'\bmod m, ~ c'\bmod m, ... , ~ h'\bmod m)$, and $a',b',c',...,h'$ are defined in $(2)$.

If there exist transformations $(5)$ converting $v_0=(3,1,1,1,0,0,0,0)$ to $v_n=(0,0,...,0)$, then for any $m\in\mathbb{N}$ must exist transformations

$$ z_1=T_m(v_0,A_1),\\ z_2=T_m(z_1,A_2),\\ ...\\ z_n=T_m(z_{n-1},A_n),\tag{5'} $$

where $z_n=(0,0,0,0,0,0,0,0)$.

---

D.

For $n=4$ ($4$ nested rdicals), if we choose $m=5,7,10,11,...$, then for $0\le A_1,A_2,A_3,A_4 <m$ value $z_4$ cannot be zero-vector.

For $n=5$ one can use $m=11,13,17,19,...$ to see that $z_5$ cannot be $\vec{0}$.

For $n=6$ one can use $m=31, 41, 43, 47, ...$ to see that $z_6$ cannot be $\vec{0}$.

So, must be at least $n\ge 7$ radicals. (I'll update this answer after checking $n=7$).

---

How D is working (temporary part):

Example for $n=5$, $m=11$:

starting vector is (a[0]=3,b[0]=1,c[0]=1,d[0]=1,e[0]=...=h[0]=0)

I need to check if there is sequence $A[0],...,A[4]$ that generates (0,0,0,0,0,0,0,0).

It can be used "as is", or more smart:
if there is sequence $A[0], ..., A[2],A[3]$ that generates vector (a,b,c,d,e,f,g,h), where $a$ can be nonzero; or even so:
if there is sequence $A[0], ..., A[2]$ that generates vector (a,b,c,d,e,f,g,h), where $a$ can be nonzero and one of $b,c,d,e,f,g,h$ is nonzero too.

If apply some modulo $m$, then each $A[j]$ is equivalent to $0$ or $1$ or $2$ or ... or $m-1$.

Then test:

for A[0]=0,1, ..., m-1 for A[1]=0,1, ..., m-1 for A[2]=0,1, ..., m-1 { a[1] = (a[0]*a[0]+ 30*h[0]*h[0] - A[0]) mod m; b[1] = 2(a[0]*b[0]+...+15*g[0]*h[0]) mod m; ... h[1] = 2(a[0]*h[0]+...+d[0]*e[0]) mod m; a[2] = (a[1]*a[1]+ 30*h[1]*h[1] - A[1]) mod m; b[2] = 2(a[1]*b[1]+...+15*g[1]*h[1]) mod m; ... h[2] = 2(a[1]*h[1]+...+d[1]*e[1]) mod m; and a[3], b[3], ..., h[3] same way; and a[4], b[4], ..., h[4] same way, but use 0 instead of A[3]; then check if only one of b[4],c[4],...,h[4] is nonzero (say d[4]); if "yes", then success (possible solution); then A[3] = a[4], and A[4] is based on this last non-zero term (like d[4]); } 

If you have something of the form $\sqrt{a+\sqrt b}$, it can be denested as follows:

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

So we have $c+d=a,cd=\frac b4$. $c$ and $d$ are therefore the solutions of $x^2-a x+\frac b4=0$. A similar process works for $\sqrt{a-\sqrt b}=\sqrt c-\sqrt d$. If the roots are rational, you have denested the square.

Similarily, for something like $\sqrt{w+\sqrt x+\sqrt y+\sqrt z}$, we have

$\sqrt a+\sqrt b + \sqrt c=\sqrt{(\sqrt a+\sqrt b + \sqrt c)^2}=\sqrt{(a+b+c)+\sqrt{4ab}+\sqrt{4ac}+\sqrt{4bc}}$, so we have $a+b+c=w,ab+ac+bc=\frac{x+y+z}{4},abc=\frac{\sqrt{xyz}}{8}.$ So $a,b,c$ are solutions of the cubic equation $u^3-wu^2+\frac{x+y+z}{4}u-\frac{\sqrt{x y z}}{8}=0$. For example,$\sqrt{10+2 \left(\sqrt{15}+\sqrt{6}+\sqrt{10}\right)}=\sqrt{10+ \sqrt{60}+\sqrt{24}+\sqrt{40}}$, has the corresponding cubic equation $u^3-10u^2+31u-30=0$, with roots $u=2,3,5$, so $\sqrt{10+2 \left(\sqrt{15}+\sqrt{6}+\sqrt{10}\right)}=\sqrt 2+\sqrt 3+\sqrt 5$.

Do you get how to do this? In general, if you have a square root with $\frac{n (n-1)}{2}$ square roots under it for some integer $n$, and exactly $1$ rational number, you can rewrite is as a sum of $n$ square roots. By squaring it and square rooting, you can equate various components, and possibly manipulating them (as done with the cubic coefficients), you can use Vieta's formulas to determine an $n$th degree polynomial whose roots determine how to denest it. For something like $\sqrt{14+\sqrt{140+\sqrt{4096+\sqrt{8847360}}}}$, you have to denest the first radical inside, then the next one... etc. You could generalise the result to higher roots than just squares, but I imagine it would be messier. As for when the denesting works, you'd better hope the roots of those polynomials are rational, or otherwise you'll end up with more nested radicals. Otherwise it doesn't have a denested form. Hope this answers your question!

The irrational number satisfies the following equation: $$x=3+\sqrt{2}+\sqrt{3}+\sqrt{5}$$ $$x^8-24 x^7+212 x^6-792 x^5+622 x^4+3768 x^3-10140 x^2+8568 x-2151=0$$ Let it be: $$x=\sqrt{a+\sqrt{b+\sqrt{c}}}$$ Now it can be shown that: $$x^8-(8a)x^7+(28a^2)x^6-(56a^3)x^5-(2b-70a^4)x^4+(8ab-56a^5)x^3-(12a^2b-28a^6)x^2+(8a^3b-8a^7)x+(b^2-2a^4b+a^8-c)=0$$ Now: $$(24-8a)x^7+(28a^2-212)x^6-(56a^3+792)x^5-(2b-70a^4-622)x^4+(8ab-56a^5-3768)x^3-(12a^2b-28a^6+10140)x^2+(8a^3b-8568-8a^7-8568)x+(b^2-2a^4b+a^8-c+2151)=0$$ Fint integral values of $a,b,c$ after putting $x=3+\sqrt2+\sqrt3+\sqrt5$.Lot of work :(.

---