# Find the least degree polynomial with integer coefficients… [closed]

Find the least degree polynomial with integer coefficients that has $\sqrt [3] {5}-\sqrt{3}$ as a zero.

## closed as off-topic by Shailesh, steven gregory, Thomas Andrews, TravisJ, Yujie ZhaJul 28 '17 at 3:29

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• Do you know the quadratic formula? Suppose, for contradiction, that there is a quadratic with integer coefficients that has $\sqrt[3]{7} - \sqrt{11}$ as a root. – Zubin Mukerjee Jul 28 '17 at 2:31
• It is similar to math.stackexchange.com/questions/574029/… but I can't find a clever substitution because of the second radical. – Aneesh Dasgupta Jul 28 '17 at 2:33
• I am new to math.stackexchange and I can't quite grasp the "enter" function as when I click it I think my cursor is moving to the next line (but I actually have to do shift+enter) hence the repetitive comments. Are members opposed to repetitive commenting like some websites or is it fine to have multiple comments in a row? Thank you. – Aneesh Dasgupta Jul 28 '17 at 2:35
• @AneeshDasgupta See Allow newline in comments. – Frenzy Li Jul 28 '17 at 2:46
• Really, stop changing your question. – Thomas Andrews Jul 28 '17 at 2:51

Hint:  let $x = \sqrt [3] {5}-\sqrt{3}\,$, then:
$$(x+\sqrt{3})^3=5 \iff x^3 + 9 x + 3 (x^2 + 1) \sqrt{3} = 5 \implies (x^3+9x-5)^2 = 27(x^2+1)^2$$
The latter gives a monic polynomial with integer coefficients of degree $6$ which has $x = \sqrt [3] {5}-\sqrt{3}\,$ as a root. What remains to be proved is that it is the minimal polynomial (which, in fact, it is)
\begin{align} x-\sqrt[3]{5}+\sqrt{3}&=0 \\ x+\sqrt{3}&=\sqrt[3]{5} \\ (x+\sqrt{3})^3&=5 \\ x^3+3\sqrt{3}x^2+9x+3\sqrt{3}&=5 \\ x^3+9x-5&=3\sqrt{3}(-x^2-1) \\ (x^3+9x-5)^2&=27(-x^2-1)^2 \\ x^6 + 18 x^4 - 10 x^3 + 81 x^2 - 90 x + 25 &= 27x^4+54x^2+27 \\ x^6 - 9 x^4 - 10 x^3 + 27 x^2 - 90 x - 2 &= 0 \end{align} We still must show that this polynomial is irreducible in order to be sure that it is the least degree polynomial with $\sqrt[3]{5}-\sqrt{3}$ as a root.
Edit: To get the irreducibility one can show that the polynomial is irreducible$\bmod 7$, or use the argument of Robert Israel given here. This polynomial takes on prime values for $$x=-611, -393, -359, -315, -263, -213, -203, -185, -107, -83, -59, -51, -21, -5, 13, 21, 39, 105, 201, 213, 313, 351, 387, 429, 507, 573$$ for a total of 26 times which means a factor of the polynomial would take on the value $\pm1$ $13$ times. It follows that the factor would take on $+1$ or $-1$ at least $7$ times which is impossible for a polynomial of degree at most $6$, hence our polynomial must be irreducible.