Polynomial Degree Help

Question

A nonzero polynomial with rational coefficients has all of the numbers$$1+\sqrt{2}, \; 2+\sqrt{3}, \;3+\sqrt{4},\; \dots, \;1000+\sqrt{1001}$$ as roots. What is the smallest possible degree of such a polynomial?

Since there are $1000$ terms, adding the radical conjugates, there will be $2000$ terms. Thus, I got the smallest possible degree would be $2000$. What am I missing here?

$3+\sqrt4=5$ and $3-\sqrt4=1$ are not conjugates. – Andreas Blass Feb 18 '20 at 00:01 — Andreas Blass, Feb 18 '20 at 00:01

Saketh Malyala · Accepted Answer · 2020-02-18T01:27:38.333

2

Not quite.

Think about $3+\sqrt{4}$: that's just $5$. A similar case holds for $8+\sqrt{9}$, etc. You would count each of those cases only once. How many such cases are there? Can you proceed?

You must count the cases from $\sqrt{2^2}$ to $\sqrt{31^2}$, for a total of $30$ cases. That means the smallest possible degree is $1970$.

edited Feb 18 '20 at 01:27

answered Feb 18 '20 at 00:01

Saketh Malyala

13,637

Ahh, I see. Thank you. I forgot that some of the numbers are whole numbers and don't have radical conjugates. – Frost Bite Feb 18 '20 at 05:21

Polynomial Degree Help

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