Let $a_0+a_1x+\dots+a_nx^n$ be a non zero polynomial with integer coefficients. If $p(√2+√3+√6)=0$, the smallest possible value of n is?

Question

Question

Let $a_0+a_1x+\dots+a_nx^n$ be a non zero polynomial with integer coefficients. If $p(√2+√3+√6)=0$, the smallest possible value of n is?

Honestly I have no idea how to begin to solve this question. And I'm guessing that the $a$'s are in arithmetic progression. Any help is appreciated.Thanks :) Options for the answers are: (A) 8; (B) 6; (C) 4; (D) 2.

Answer 1

Given the question is about a multiple choice test with possible answers $2,4,6,8$ maybe the most direct and elementary proof is to notice that, with $x=\sqrt 2 + \sqrt 3 + \sqrt 6\;$:

• $(x-\sqrt 6)^2=(\sqrt 2 + \sqrt 3)^2 \iff x^2+6-2x \sqrt 6 = 2 + 3 + 2 \sqrt 6\;$ then isolating $\sqrt 6$ on one side and squaring again we get a $4^{th}$ degree equation in $x$, which eliminates answers $n=6,8$.

• To eliminate $n=2$, note that $x^2=11 + 6 \sqrt 2 + 4 \sqrt 3 + 2 \sqrt6$. If an equation existed with rational coefficients $a x^2 + b x + c = 0$ then that would give a linear dependency between $\sqrt 2, \sqrt 3,\sqrt6$, but they are linearly independent over the rationals (see for example How to prove $1$,$\sqrt{2},\sqrt{3}$ and $\sqrt{6}$ are linearly independent over $\mathbb{Q}$?).

Which leaves the answer $n=4$.

Answer 2

$\alpha=\sqrt 2+\sqrt 3+\sqrt 6 \in \mathbb Q[\sqrt 2,\sqrt 3]$, which has degree $4$ over $\mathbb Q$. A basis is $1,\sqrt 2,\sqrt 3,\sqrt 6$.

So, the degree of $\alpha$ is at most $4$ and must be divisor of $4$.

The degree is not $1$, because $\alpha$ is not rational, since otherwise $1,\sqrt 2,\sqrt 3,\sqrt 6$ would be linearly dependent.

The degree is not $2$, because $\alpha \notin \mathbb Q[\sqrt 2]$ and $\alpha \notin \mathbb Q[\sqrt 3]$, since otherwise $1,\sqrt 2,\sqrt 3,\sqrt 6$ would be linearly dependent.

So, the only possibility left is that the degree is $4$.

Answer 3

Let's delve into the algebra behind @piquito's answer to see where the decreased degree comes from when one radical is the product of the other two.

Start with $x=\sqrt{a}+\sqrt{b}+\sqrt{c}$, with $a, b, c$ all positive and rational but none of them a rational square. A polynomial having this as a root must contain the factor

$x-\sqrt{a}-\sqrt{b}-\sqrt{c}$.

We incorporate this factor into the following fourth-degree product in which the signs on the radicals form an orthgonal array:

$P(x)=$

$(x-\sqrt{a}-\sqrt{b}-\sqrt{c})·$

$(x+\sqrt{a}+\sqrt{b}-\sqrt{c})·$

$(x+\sqrt{a}-\sqrt{b}+\sqrt{c})·$

$(x-\sqrt{a}+\sqrt{b}+\sqrt{c})$.

With such an orthogonal array of signs we might expect that when we multiply the factors out all the radicals will cancel out because a positive radical term is balanced by a corresponding negative one. In fact such a prediction is almost correct:

$P(x)=x^4-2(a+b+c)x^2-8\sqrt{abc}x+2(ab+ac+bc)-(a^2+b^2+c^2)$.

There is one surviving radical term symmetric in $a, b, c$. To clear it we multiply $P(x)$ as defined above by $P(-x)$; we have to incorporate the sign of $x$ itself into the orthogonal array which is thereby doubled in size. That product is the usual eighth-degree minimal polynomial.

But suppose that in the above development $c=r^2ab$ where $r$ is any positive rational number (it does not have to be $1$). Then $\sqrt{abc}=rab$ and now our fourth-degree polynomial has only rational coefficients:

$P(x)=x^4-2(a+b+c)x^2-8(rab)x+2(ab+ac+bc)-(a^2+b^2+c^2)$.

Thereby the minimal polynomial degree is reduced from eight to four.

Answer 4

One approach to solve this without any advanced theorems is to start with a guess like:

$$P(x) = A_0 + A_1x^1 + A_2x^2 + A_3x^3 + A_4x^4$$

Then expand

$$P(\sqrt{2} + \sqrt{3} + \sqrt{6}) = 0$$

and you'll some equation of the form:

$$B_0 + B_1 \sqrt{2} + B_2 \sqrt{3} + B_3 \sqrt{6} = 0$$

where none of the $B$ contain any radicals, they are just linear combinations of the $A_k$. In this case it is that:

$$\begin{align} B_0 &= A_0 + 11~A_2 + 36~A_3 + 265~A_4 \\ B_1 &= A_1 + 6~A_2 + 29~A_3 + 180~A_4 \\ B_2 &= A_1 + 4~A_2 + 27~A_3 + 136~A_4 \\ B_3 &= A_1 + 2~A_2 + 21~A_3 + 92~A_4 \\ \end{align}$$

If you take it for granted, that there are no integers $m, n$ such that $m\sqrt{x} + n\sqrt{y} = 0$ except for $m = 0$ and $n = 0$ (for sufficiently different $x$, $y$), then since $P(x) = 0$ then

$$\begin{align} 0 &= A_0 + 11~A_2 + 36~A_3 + 265~A_4 \\ 0 &= A_1 + 6~A_2 + 29~A_3 + 180~A_4 \\ 0 &= A_1 + 4~A_2 + 27~A_3 + 136~A_4 \\ 0 &= A_1 + 2~A_2 + 21~A_3 + 92~A_4 \\ \end{align}$$

That's 4 equations and 5 variables. Naturally you'll have an extra variable, since if $P(x)$ is a solution, then so is $n~P(x)$. Since the question only asked if such a polynomial exists, you could have skipped most of the calculations and just pointed out that such calculations were possible without actually doing them. If you do bother to do the calculations, the result is $P(x) = n(23 + 48~x + 22x^2 + x^4)$ for any integer $n$.

Also, the reason there are only 4 values in the $B$ array is because $\sqrt{2}\sqrt{3} = \sqrt{6}$, so that limits the number of radical coefficients after the expansion of $P(\sqrt{2} + \sqrt{3} + \sqrt{6})$. If we had expanded something like $P(\sqrt{2} + \sqrt{3} + \sqrt{11})$, then there would have been $2^3$ instead of $2^2$ terms in the expansion.

Answer 5

The degree is $4$ because $\sqrt6=\sqrt2\cdot\sqrt3$ (in general the degree for the root $\sqrt a+\sqrt b+\sqrt c$ is $8$ but the relation $\sqrt6=\sqrt2\cdot\sqrt3$ makes important algebraic restriction which gives degree $4$).

The calculation of the minimal polynomial is straightforward giving

$$x^4-22x^2-48x-23=0$$ The roots are $$x_1=\sqrt2+\sqrt 3+\sqrt 6\approx5.5958\\x_2\approx-2.7673\\x_3\approx-2.1317\\x_4\approx -0.6967$$