Find a polynomial $p$ with integer coefficients for which $a = \sqrt{2} + \sqrt[3]{2}$ is a root. That is find $p$ such that for some non-negative integer $n$, and integers $a_0$, $a_1$, $a_2$, ..., $a_n$, $p(x) = a_0 + a_1 x + a_2 x^2 + ... + a_n x^n$, and $p(a) = 0$.

I do not know how to solve this. It is very challenging. Also, if you name any theorem please describe it in a way that is easy to understand. If you just name it, I won't be able to understand it. (My math might not be/is not as good as yours.)

Thanks for any help!

  • $\begingroup$ Why has the post been deleted? $\endgroup$ – James Moriarty Mar 10 '16 at 10:06
  • $\begingroup$ There was an arithmetical error. $\endgroup$ – J.R. Mar 10 '16 at 10:06
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    $\begingroup$ But it can be fixed. It might be easier to start with $2 = (a - \sqrt 2)^3$ ... $\endgroup$ – Martin R Mar 10 '16 at 10:07
  • $\begingroup$ Was that arithmetic error in squaring? a^2 = 2+2.(2^(1/6))+(2^(2/3)), while you had entered 2+4.(2^(1/6))+(2)^(2/3). Apart from that I don't think there was any other arithmetic error. $\endgroup$ – James Moriarty Mar 10 '16 at 10:21
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    $\begingroup$ This is from SUMAC2016 qualifying exam. If you want to cheat in that exam or support cheating, ... $\endgroup$ – Jyrki Lahtonen Mar 10 '16 at 18:13

We have $$ a=\sqrt2+\sqrt[3]2\\ a-\sqrt2=\sqrt[3]2\\ (a-\sqrt2)^3=2\\ a^3-3\sqrt2a^2+6a-2\sqrt2=2\\ a^3+6a-2=\sqrt2(3a^2+2)\\ (a^3+6a-2)^2=2(3a^2+2)^2 $$ which has all integer coefficients, once you expand the brackets.

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    $\begingroup$ Expanded form: a^6-6a^4-4a^3+12a^2-24a-4=0 $\endgroup$ – James Moriarty Mar 10 '16 at 10:49
  • $\begingroup$ @JamesMoriarty Yes, I suppose it is. I am on mobile right now, so I didn't think I would be able to do the last squatting and subsequent tidying up correctly. $\endgroup$ – Arthur Mar 10 '16 at 11:03

There is a simple way to do this. Consider a matrix $M_1$ whose eigenvalues are $\pm \sqrt{2},$ and a matrix $M_2,$ whose eigenvalues are the roots of $x^3-2$ - this is easy to do by the companion matrix construction. Now, consider the matrix $M_1 \otimes I_3 + I_2 \otimes M_2 ,$ where $I_k$ is the $k\times k$ identity matrix. Its characteristic polynomial is what you are looking for.


Let a row vector, $r$, represent a real number by $$ r\, \begin{bmatrix} 1\\2^{1/6}\\2^{2/6}\\2^{3/6}\\2^{4/6}\\2^{5/6} \end{bmatrix} $$ multiplication of the row vector $r$ by $2^{1/3}+2^{1/2}$ can represented as $rM$ where $$ M= \begin{bmatrix} 0&0&1&1&0&0\\ 0&0&0&1&1&0\\ 0&0&0&0&1&1\\ 2&0&0&0&0&1\\ 2&2&0&0&0&0\\ 0&2&2&0&0&0\\ \end{bmatrix} $$ The top row of $M^k$ gives the representation of $\left(2^{1/3}+2^{1/2}\right)^k$. If $P$ is the characteristic polynomial of $M$, then $P(M)=0$. Thus, we also have $P\!\left(2^{1/3}+2^{1/2}\right)=0$, since that is represented by the top row of $P(M)$.

The characteristic polynomial of $M$ is $$ P(x)=x^6-6x^4-4x^3+12x^2-24x-4 $$ Therefore, $$ P\!\left(2^{1/3}+2^{1/2}\right)=0 $$

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    $\begingroup$ would the downvoter care to comment? $\endgroup$ – robjohn Mar 10 '16 at 14:11

Here's a general approach to this kind of problem.

[NOTE: I misread the question as involving $\sqrt{2}$ and $\sqrt[3]{a}$ instead of $\sqrt{2}$ and $\sqrt[3]{2}$, whose sum is being called $a$. This means that some of the specific algebraic calculations that follow aren't exactly the ones the original questioner needs, but the overall structure is the same whatever you're taking the cube root of so I'll leave the mistake in. You could just put $a=2$ in what I wrote to get an answer to the original question.]

Consider those roots -- the square root of 2, and the cube root of $a$. There are actually two square roots of 2 (which we usually write as $\sqrt{2}$ and $-\sqrt{2}$, but as far as equation-solving goes they're interchangeable). And there are three cube roots of $a$; if we write $\omega=(-1+i\sqrt{3})/2$ they are $\sqrt[3]{a},\omega\sqrt[3]{a},\omega^2\sqrt[3]{a}$. (The point is that $1,\omega,\omega^2$ are the things whose cube is 1, just as $1,-1$ are the things whose square is 1.)

If we have an integer polynomial with, $\sqrt{2}+\sqrt[3]{a}$ as a root, then it "can't tell the difference" between that and, say, $-\sqrt{2}+\omega^2\sqrt[3]{a}$ so that had better be a root as well. (If you want more details, the technical term to look up is "conjugate" or, more specfically, "Galois conjugate".)

Therefore, an integer polynomial with $\sqrt{2}+\sqrt[3]{a}$ as a root must have all six of these "conjugates" as roots. So we have


and now expanding this out gives the polynomial we need.

In practice it may be easier first to consider the cubic polynomial


which turns out, if I've done my algebra right, to equal $x^3-3\sqrt{2}x^2+6x-2\sqrt2-a$. And then we multiply this by its conjugate (with the signs of all the $\sqrt2$s flipped) to get the degree-6 polynomial we need.

So the general recipe is: take the root you need, find all its conjugates, and multiply together all the corresponding $(x-\textrm{root})$ factors. The resulting polynomial's coefficients will all come out rational.


A systematic way:

Compute the first powers of $a$, keeping the fractional powers of $2$ under $1$:

$$a=2^{1/2}+2^{1/3}\\ a^2=2+2\cdot2^{5/6}+2^{2/3}\\ a^3=2\cdot2^{1/2}+6\cdot2^{1/3}+6\cdot2^{1/6}+2\\ \cdots$$

You are forming a linear system of equations in the five "unknowns" $2^{1/6},2^{1/3},2^{1/2},2^{2/3},2^{5/6}$. So continuing to the sixth degree you will end up with an overdetermined system, and you can find a linear combination of the LHS with integer coefficients such that the RHS is an integer.

I guess that this method, even if it is tedious, works with any linear combination of fractional powers.


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