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What is the minimum degree of an equation with rational coefficients that has a root $x=a+\sqrt{b}+\sqrt{c}+\sqrt{d}$ with $a,b,c,d$ primes numbers?

I know how to find an equation of second degree that has root $a+\sqrt{b}$ $$ x=a+\sqrt{b} \quad \rightarrow \quad (x-a)^2=b $$

and a $4-$degree equation that has root $a+\sqrt{b}+\sqrt{c}$ $$ x=a+\sqrt{b}+\sqrt{c} \quad \rightarrow \quad (x-a)^2=(\sqrt{b}+\sqrt{c})^2 \quad \rightarrow \quad \left[(x-a)^2-b-c \right]^2=4bc $$

But it seems that this simple method cannot be used for a root with more than two surds. There is it some other method?

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    $\begingroup$ It will be degree eight. This is a primitive element of the field extension $\Bbb{Q}(\sqrt b,\sqrt c,\sqrt d)$ over $\Bbb{Q}$. The question is also a special case of this older question. See here for a proof of the fact that the above field extension is of degree eight. $\endgroup$ Apr 14 '17 at 20:44
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    $\begingroup$ Anyway, the minimal polynomial has zeros $a\pm \sqrt b\pm\sqrt c\pm\sqrt d$. All sign combinations occur. $\endgroup$ Apr 14 '17 at 20:48
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Let $x=a+ \sqrt{b}+ \sqrt{c}+ \sqrt{d}$ square this and move some terms around \begin{eqnarray*} (x-a)^2+b-c-d = -2(x-a)\sqrt{b}+ 2\sqrt{c} \sqrt{d} \end{eqnarray*} Square it again and move some more terms around \begin{eqnarray*} ((x-a)^2+b-c-d)^2 -4b(x-a)^2-4cd = -8(x-a)\sqrt{b}\sqrt{c} \sqrt{d} \end{eqnarray*} Squaring one final time & we have \begin{eqnarray*} (((x-a)^2+b-c-d)^2 -4b(x-a)^2-4cd)^2 = 64(x-a)^2bcd \end{eqnarray*} So the equation that this quantity satisfies an equation of degree $\color{red}{8}$ as expected.

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    $\begingroup$ How will you prove that this is minimum degree polynomial with that required root? $\endgroup$ Apr 14 '17 at 21:03
  • $\begingroup$ @JaideepKhare I have shown that $8$ is an upper bound on the degree & there are probably situtations where a lower value will work ... E.g. $d=bc$ ... I am not sure at the moment to make my answer completely sound ... You are right there is still something left to prove. $\endgroup$ Apr 14 '17 at 21:23
  • $\begingroup$ @JaideepKhare That is a bit delicate. If $b,c,d$ are distinct primes (what was probably intended), it turns out that 8 is the minimum degree. You need the machinery of field extensions (possibly also Galois theory) to prove this. See the questions I linked to for the arguments. Took me a while to check the end result. Hence the belated +1 $\endgroup$ Apr 14 '17 at 21:27
  • $\begingroup$ @JyrkiLahtonen Than that's beyond my scope.(Presently, because I am a highschool student) $\endgroup$ Apr 14 '17 at 21:34
  • $\begingroup$ @JaideepKhare It is not out of the question that you would get at least the rough idea from a suitable book. WP 1,2 may be too sketchy. $\endgroup$ Apr 14 '17 at 21:40

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