Polynomial Roots Set Size
A set $ S $ is constructed as follows. To begin, $ S = \{0,10\} $. Repeatedly, as long as possible, if $ x $ is an integer root of some nonzero polynomial $ a_{n}x^n + a_{n-1}x^{n-1} + \dots + a_{1}x + a_0 $ for some $ n\geq{1} $, all of whose coefficients $ a_i $ are elements of $ S $, then $ x $ is put into $ S $. When no more elements can be added to $ S $, how many elements does $ S $ have?
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
We start with the set $ S = \{0,10\} $. We can construct the polynomial $ 10x + 10 = 0, $ which has $ x = -1 $ as a root. Thus, we can expand our set to $ S = \{-1,0,10\} $.
We can then construct the polynomial
\[10x^{10} - x^9 - x^8 - x^7 - x^6 - x^5 - x^4 - x^3 - x^2 - x - 1 = 0,\]which has $ x = 1 $ as a root, and we can construct the polynomial $ -x^3 - x + 10 = 0, $ which has $ x = 2 $ as a root. Thus, we can expand our set to $ S = \{-1, 0, 1, 2, 10\} $.
Next, we can construct the polynomial $ x + 10 = 0, $ which has $ x = -10 $ as a root, the polynomial $ 2x + 10 = 0, $ which has $ x = -5 $ as a root, and the polynomial $ x + 2 = 0, $ which has $ x = -2 $ as a root. Our set $ S $ is now $ \{-10, -5, -2, -1, 0, 1, 2, 10\} $.
Finally, we can construct the polynomial $ x - 5 = 0, $ which has $ x = 5 $ as a root, giving us the set
\[S = \{-10, -5, -2, -1, 0, 1, 2, 5, 10\}.\]Now, suppose we construct the polynomial
\[a_n x^n + a_{n - 1} x^{n - 1} + \dots + a_1 x + a_0 = 0,\]with coefficients from the set $ S = \{-10, -5, -2, -1, 0, 1, 2, 5, 10\} $. If $ a_0 = 0, $ then we can factor out some power of $ x, $ to obtain a polynomial where the constant term is nonzero. Thus, we can assume that $ a_0 \neq 0 $.
By the Integer Root Theorem, any integer root of this polynomial must divide $ a_0 $. But we see that any divisor of a non-zero element in $ S $ already lies in $ S, $ so we cannot expand the set $ S $ any further. Hence, the answer is $ \boxed{9} $ elements.
Freeflows