Least degree polynomial with integer coefficient and having one root $\sqrt{8+\sqrt 6 +4\sqrt 3+ 3\sqrt 2}$

Question

Find Least degree polynomial with integer coefficient and having one root as $\sqrt{8+\sqrt 6 +4\sqrt 3+ 3\sqrt 2}$

I tried doing square again and again is there any simpler method

Answer 1

Let us ask a computer algebra system, sage, about this:

sage: a = sqrt(8+sqrt(6)+4*sqrt(3)+3*sqrt(2))
sage: a.minpoly()
x^8 - 32*x^6 + 240*x^4 - 320*x^2 - 368

So the "blind result" is telling us that we have to "square three times". The first squaring is simple, so let us consider $b={8+\sqrt 6 +4\sqrt 3+ 3\sqrt 2}$ instead of the above $a=\sqrt{8+\sqrt 6 +4\sqrt 3+ 3\sqrt 2}$. Then a possibility to do the job as a human is to consider the Galois conjugates of $b$, working in the field $\Bbb Q(\sqrt 2,\sqrt 3)$ where $b$ lives in: $$ \begin{aligned} b=b_{00} &=8+\sqrt 6 +4\sqrt 3+ 3\sqrt 2\ ,\\ b_{10} &=8-\sqrt 6 +4\sqrt 3- 3\sqrt 2\ ,\\ b_{01} &=8-\sqrt 6 -4\sqrt 3+ 3\sqrt 2\ ,\\ b_{11} &=8+\sqrt 6 -4\sqrt 3- 3\sqrt 2\ , \end{aligned} $$ and building the polynomial $(X-b_{00})(X-b_{10})(X-b_{01})(X-b_{11})$. Now it is easy to see the coefficient in $X^3$, which is $-\sum b_{jk}=-(8+8+8+8)=-32$. The free coefficient can by also computed in a few lines, $$ \begin{aligned} &b_{00}b_{10}b_{01}b_{11} \\ &\qquad=(b_{00}b_{10})\cdot(b_{01}b_{11}) \\ &\qquad= \Big(\ (8+4\sqrt 3)^2 - (\sqrt 6 + 3\sqrt 2)^2\ ) \Big(\ (8-4\sqrt 3)^2 - (\sqrt 6 - 3\sqrt 2)^2\ ) \\ &\qquad= \Big(\ (112+64\sqrt 3) - (24+12\sqrt 3)\ ) \Big(\ (112-64\sqrt 3) - (24-12\sqrt 3)\ ) \\ &\qquad= \Big(\ 88+52\sqrt 3\ ) \Big(\ 88-52\sqrt 3\ ) \\ &\qquad=-368\ . \end{aligned} $$ Yes, we have a match of the free coefficient.

Let us do the job also for the coefficient of $X^2$ in the minimal polynomial of $b$, it is $$ \begin{aligned} &b_{00}b_{10}+(b_{00}+b_{10})\cdot(b_{01}+b_{11})+b_{01}b_{11} \\ &\qquad= (88+52\sqrt 3)+2(8+4\sqrt 3)\cdot 2(8+4\sqrt 3)+(88-52\sqrt 3) \\ &\qquad= 88+88+4(64-48) \\ &\qquad= 240 \ . \end{aligned} $$ The last coefficient can be computed along the same lines, grouping $b_{00}b_{10}\cdot(b_{01}+b_{11})+ (b_{00}+b_{10})\cdot b_{01}b_{11}$, and the half of the computations are already done.

Note: The indices $jk$ have the meaning $$ b_{jk}= 8 + (-1)^j 3\sqrt 2 +(-1)^k 4\sqrt 3+ (-1)^j \sqrt 2\cdot (-1)^k \sqrt 3 \ , $$ corresponding to the Galois automorphism sending $\sqrt 2\to(-1)^j\sqrt 2$ and $\sqrt 3\to(-1)^k \sqrt 3$.