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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?

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    $\begingroup$ $3+\sqrt4=5$ and $3-\sqrt4=1$ are not conjugates. $\endgroup$ Feb 18, 2020 at 0:01

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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 $1970$.

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  • $\begingroup$ Ahh, I see. Thank you. I forgot that some of the numbers are whole numbers and don't have radical conjugates. $\endgroup$
    – Frost Bite
    Feb 18, 2020 at 5:21

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