Annihilator polynomial of $\sqrt{5} + \sqrt[5]{2}$

Question

I can find an annihilator polynomial of $\alpha = \sqrt{5} + \sqrt[5]{2}$.

But I have an exercise which asks me to show that $$P = X^{10} - 25 X^8 + 250 X^6 - 4X^5 - 1250X^4 - 200 X^3 + 3125 X^2 - 500X - 3121$$ is also an annihilator polynomial.

How can I do it simply ? I could compute $P(\alpha)$, but this is not very easy...

The minimal polynomial is of degree 10, so finding it suffices (it is unique, if we normalize it to be a monic polynomial). Alternatively you can try and put it into the form on the r.h.s. of Math Gems' answer, but that smacks of reverse-engineering - not always a bad thing! — Jyrki Lahtonen, Apr 08 '13 at 04:16
Nothing stupid. Inexperienced, may be, but time will fix that. — Jyrki Lahtonen, Apr 08 '13 at 07:00

score 4 · Accepted Answer · answered Apr 07 '13 at 12:33

4

Hint: $(\alpha-\sqrt5)^5=2$. Expand the l.h.s. using the binomial formula (row five from Pascal's triangle). Move the terms containing $\sqrt5$ to one side. Square. Expand again.

answered Apr 07 '13 at 12:33

Jyrki Lahtonen

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Math Gems · Answer 2 · 2013-04-07T15:23:33.063

1

$\rm x=\sqrt{a}\!+\!\sqrt[5]{b}\:\Rightarrow\:0=((x\!-\!\sqrt{a})^5\!-\!b)((x\!+\!\sqrt{a})^5\!-b) = (x^2\!-\!a)^5\!-2b(x^5\!+\!10ax^3\!+\!5a^2x)\!+\!b^2$

edited Apr 07 '13 at 15:23

answered Apr 07 '13 at 15:18

Math Gems

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Annihilator polynomial of $\sqrt{5} + \sqrt[5]{2}$

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