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Final Minus Two (Posted on 2015-01-24) Difficulty: 3 of 5
Start with the set S = {0, 2014}. Then, repeatedly, expand S as follows.

Place into S any integer that is a root of a polynomial the coefficients of which are in S.

Prove that the negative number −2 eventually appears in S.

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution What am I missing this time (spoiler?) | Comment 1 of 2
x = -2 is a root of the polynomial x + 2 = 0.

So, it seems like x = -2 gets added during the first expansion of S.

I must be missing something, if this problem is difficulty 3.


Oh, now I see.  S starts with only two elements, 0 and 2014.  Back to the drawing board.

Edited on January 24, 2015, 1:57 pm
  Posted by Steve Herman on 2015-01-24 13:52:27

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