Find integers $a,b,c,d$ such that
is it possible for someone to run a computer search for me to turn up integers $a,b,c,d$ such that $$1+\sqrt{2}+\sqrt{3}+\sqrt{6}=\sqrt{a+\sqrt{b+\sqrt{c+\sqrt{d}}}}$$? Note: I know they exist, I just can't find them.
Solution 1:
Let $s = 1 + \sqrt{2}+\sqrt{3}+\sqrt{6}$, so you want $s^2 = a + \sqrt{b+\sqrt{c+\sqrt{d}}}$.
Now $s^2$ has minimal polynomial $p(z) = z^4-48 z^3+200 z^2-192 z+16$ over the rationals, of degree $4$. On the other hand, $a + \sqrt{b+\sqrt{c+\sqrt{d}}}$ satisfies
$q(z) = (((z-a)^2 -b)^2 - c)^2 - d = 0$, a polyunomial of degree $8$. So we need $q(z)$ to be divisible by $p(z)$. The remainder of $q(z)$ on division by $p(z)$ is
$$ r(z) = \left( -56\,{a}^{5}+3360\,{a}^{4}+80\,{a}^{3}b-117824\,{a}^{3}-2880\,
{a}^{2}b-24\,a{b}^{2}+2564352\,{a}^{2}+50496\,ab+8\,ac+288\,{b}^{2}-
31875456\,a-366336\,b-96\,c+173339136 \right) {z}^{3}+ \left( 28\,{a}^
{6}-60\,{a}^{4}b-14000\,{a}^{4}+36\,{a}^{2}{b}^{2}+526848\,{a}^{3}+
12000\,{a}^{2}b-12\,{a}^{2}c-4\,{b}^{3}-11524800\,{a}^{2}-225792\,ab-
1200\,{b}^{2}+4\,bc+143308800\,a+1646400\,b+400\,c-779335936 \right) {
z}^{2}+ \left( -8\,{a}^{7}+24\,{a}^{5}b-24\,{a}^{3}{b}^{2}+13440\,{a}^
{4}+8\,{a}^{3}c+8\,a{b}^{3}-515200\,{a}^{3}-11520\,{a}^{2}b-8\,abc+
11289600\,{a}^{2}+220800\,ab+1152\,{b}^{2}-140403712\,a-1612800\,b-384
\,c+763545600 \right) z+{a}^{8}-4\,{a}^{6}b+6\,{a}^{4}{b}^{2}-2\,{a}^{
4}c-4\,{a}^{2}{b}^{3}-1120\,{a}^{4}+4\,{a}^{2}bc+{b}^{4}+43008\,{a}^{3
}+960\,{a}^{2}b-2\,{b}^{2}c-942592\,{a}^{2}-18432\,ab-96\,{b}^{2}+{c}^
{2}+11722752\,a+134656\,b+32\,c-d-63750912
$$
so we look for integer solutions of the system of equations setting the
coefficients of $r(z)$ to $0$. Using Maple I took a "$\text{plex}(d,c,b,a)$'' Groebner basis of this system. The first basis element factors as
$$ \left( a-24 \right) \left( 3\,a-68 \right) \left( a-21 \right)
\left( {a}^{2}-24\,a+36 \right) \left( {a}^{2}-24\,a+48 \right)
\left( {a}^{2}-24\,a+16 \right)
$$
so if $a$ is an integer it can only be $21$ or $24$.
With $a=21$ we get $b = 413$, $c=4656$, $d=16588800$, and indeed (again, according to Maple)
$s = \sqrt{21 + \sqrt{413+\sqrt{4656+\sqrt{16588800}}}}$.
Solution 2:
Problem statement: $$1+\sqrt{2}+\sqrt{3}+\sqrt{6}=\sqrt{a+\sqrt{b+\sqrt{c+\sqrt{d}}}}$$ Let's try this identity that you posted: $$\sqrt{x}+\sqrt{y}=\sqrt{x+y+\sqrt{4xy}}$$ Use your variables: $$\sqrt{3}+\sqrt{6}$$ $$=\sqrt{3+6+\sqrt{4\cdot 3\cdot 6}}$$ $$=\sqrt{9+\sqrt{72}}$$ $$\sqrt{1}+\sqrt{2}$$ $$=\sqrt{1+2+\sqrt{4\cdot 1 \cdot 2}}$$ $$=\sqrt{3+\sqrt{8}}$$ $$\sqrt{9+\sqrt{72}}+\sqrt{3+\sqrt{8}}$$ $$=\sqrt{\left(9+\sqrt{72}\right)+\left(3+\sqrt{8}\right)+\sqrt{4\cdot \left(9+\sqrt{72}\right)\cdot\left(3+\sqrt{8}\right)}}$$ $$=\sqrt{12+\sqrt{72}+\sqrt{8}+\sqrt{\left(36+\sqrt{1152}\right)\cdot\left(3+\sqrt{8}\right)}}$$ $$=\sqrt{12+\sqrt{72}+\sqrt{8}+\sqrt{108+3\sqrt{1152}+36\sqrt{8}+\sqrt{9216}}}$$ $$=\sqrt{12+8\sqrt{2}+\sqrt{108+144\sqrt{2}+\sqrt{9216}}}$$ $$=\sqrt{12+8\sqrt{2}+\sqrt{108+144\sqrt{2}+96}}$$ $$=\sqrt{12+8\sqrt{2}+\sqrt{204+144\sqrt{2}}}$$ $$=\sqrt{12+\sqrt{128}+\sqrt{204+\sqrt{41472}}}$$ Almost there!