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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?

 May 4, 2019
 #1
avatar+5788 
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\(\text{The polynomial evaluated at 1 is the sum of the coefficients. This must be a rational number}\\ \text{Thus each root that is not a rational number must have it's conjugate also as a root}\\ \text{This will be all the roots of the form }a + \sqrt{b} \text{ where }\sqrt{b} \not \in \{2,3,\dots,1001\}\)

 

\(\text{There are 30 numbers in 2-1001 that are perfect squares}\\ \text{So }1000-30=970 \text{ of the roots must also have their conjugate as a root}\\ \text{Thus we have a total of }\\ 30 + 2\cdot 970 = 1970 \text{ is the smallest possible degree of the polynomial}\)

 

I'm pretty sure this is correct but I might be missing something.  Someone else should take a look.

 May 4, 2019
 #2
avatar+28135 
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Your reasoning looks good to me!

Alan  May 5, 2019
 #3
avatar+133 
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Thanks I understand now

 

It took me awhile

 May 12, 2019

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