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.
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.
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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
What am I missing this time (spoiler?)
