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?

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.