Problem:
A set is constructed as follows. To begin, . Repeatedly, as long as possible, if is an integer root of some polynomial for some , all of whose coefficients are elements of , then is put into . When no more elements can be added to , how many elements does have?
Answer Choices:
A.
B.
C.
D.
E.
Solution:
Because -1 is a root of is added to . Then 1 is also added to , because it is a root of . At this point -10 , a root of , can be added to . Because 2 is a root of , and -2 is a root of , both 2 and -2 can be added to . The polynomials and allow 5 and -5 into . At this point . No more elements can be added to , because by the Rational Root Theorem, any integer root of a polynomial with integer coefficients whose constant term is a factor of 10 must be a factor of 10 . Therefore contains elements.
Note: It is not true that in general if starts with then all factors of can be added to . For example, applying the procedure to gives only , although of course it takes some argument to rule out ±5 and ±7 .
The problems on this page are the property of the MAA's American Mathematics Competitions