# 2012 AMC 12B Problems/Problem 23

## Problem 23

Consider all polynomials of a complex variable, $P(z)=4z^4+az^3+bz^2+cz+d$, where $a,b,c,$ and $d$ are integers, $0\le d\le c\le b\le a\le 4$, and the polynomial has a zero $z_0$ with $|z_0|=1.$ What is the sum of all values $P(1)$ over all the polynomials with these properties? $\textbf{(A)}\ 84\qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 108\qquad\textbf{(E)}\ 120$

## Solution

First, assume that $z_0\in \mathbb{R}$, so $z_0=1$ or $-1$. $1$ does not work because $P(1)\geq 4$. Assume that $z_0=-1$. Then $0=P(-1)=4-a+b-c+d$, we have $4+b+d=a+c\leq 4+b$, so $d=0. Also,$a=4 $has to be true since$4+b=a+c \leq a+b $. Now$4+b=4+c $deduces$b=c $, therefore the only possible choices for$(a,b,c,d) $are$(4,t,t,0) $. In these cases,$P(-1)=4-4+t-t+0=0 $. The sum of$P(1) $over these cases is$\sum_{t=0}^{4} (4+4+t+t) = 40+20=60$. Second, assume that$ (Error compiling LaTeX. ! Missing $inserted.)z_0\in \mathbb{C} \backslash \mathbb{R}$.

==Solution (doubtful) ==

Since$(Error compiling LaTeX. ! Missing$ inserted.)z_0 $is a root of$P $, and$P $has integer coefficients,$z_0 $must be algebraic. Since$z_0 $is algebraic and lies on the unit circle,$z_0 $must be a root of unity (Comment: this is not true. See this link: [http://math.stackexchange.com/questions/4323/are-all-algebraic-integers-with-absolute-value-1-roots-of-unity]). Since$P $has degree 4, it seems reasonable (and we will assume this only temporarily) that$z_0 $must be a 2nd, 3rd, or 4th root of unity. These are among the set$\{\pm1,\pm i,(-1\pm i\sqrt{3})/2\} $. Since complex roots of polynomials come in conjugate pairs, we have that$P $has one (or more) of the following factors:$z+1 $,$z-1 $,$z^2+1 $, or$z^2+z+1 $. If$z=1 $then$a+b+c+d+4=0 $; a contradiction since$a,b,c,d $are non-negative. On the other hand, suppose$z=-1 $. Then$(a+c)-(b+d)=4 $. This implies$a+b=8,7,6,5,4 $while$b+d=4,3,2,1,0 $correspondingly. After listing cases, the only such valid$a,b,c,d $are$4,4,4,0 $,$4,3,3,0 $,$4,2,2,0 $,$4,1,1,0 $, and$4,0,0,0$. Now suppose$ (Error compiling LaTeX. ! Missing $inserted.)z=i $. Then$4=(a-c)i+(b-d) $whereupon$a=c $and$b-d=4 $. But then$a=b=c $and$d=a-4 $. This gives only the cases$a,b,c,d $equals$4,4,4,0$, which we have already counted in a previous case.

Suppose$(Error compiling LaTeX. ! Missing$ inserted.)z=-i $. Then$4=i(c-a)+(b-d) $so that$a=c $and$b=4+d $. This only gives rise to$a,b,c,d $equal$4,4,4,0$which we have previously counted. Finally suppose$ (Error compiling LaTeX. ! Missing $inserted.)z^2+z+1 $divides$P $. Using polynomial division ((or that$z^3=1 $to make the same deductions) we ultimately obtain that$b=4+c $. This can only happen if$a,b,c,d $is$4,4,0,0$.

Hence we've the polynomials <cmath>4x^4+4x^3+4x^2+4x</cmath> <cmath>4x^4+4x^3+3x^2+3x</cmath> <cmath>4x^4+4x^3+2x^2+2x</cmath> <cmath>4x^4+4x^3+x^2+x</cmath> <cmath>4x^4+4x^3</cmath> <cmath>4x^4+4x^3+4x^2</cmath> However, by inspection$(Error compiling LaTeX. ! Missing$ inserted.)4x^4+4x^3+4x^2+4x+4 $has roots on the unit circle, because$x^4+x^3+x^2+x+1=(x^5-1)/(x-1) $which brings the sum to 92 (choice B). Note that this polynomial has a 5th root of unity as a root. We will show that we were \textit{almost} correct in our initial assumption; that is that$z_0 $is at most a 5th root of unity, and that the last polynomial we obtained is the last polynomial with the given properties. Suppose that$z_0 $in an$n $th root of unity where$n>5 $, and$z_0 $is not a 3rd or 4th root of unity. (Note that 1st and 2nd roots of unity are themselves 4th roots of unity). If$n $is prime, then \textit{every}$n $th root of unity except 1 must satisfy our polynomial, but since$n>5 $and the degree of our polynomial is 4, this is impossible. Suppose$n $is composite. If it has a prime factor$p $greater than 5 then again every$p $th root of unity must satisfy our polynomial and we arrive at the same contradiction. Therefore suppose$n $is divisible only by 2,3,or 5. Since by hypothesis$z_0 $is not a 2nd or 3rd root of unity,$z_0 $must be a 5th root of unity. Since 5 is prime, every 5th root of unity except 1 must satisfy our polynomial. That is, the other 4 complex 5th roots of unity must satisfy$P(z_0)=0 $. But$(x^5-1)/(x-1) $has exactly all 5th roots of unity excluding 1, and$(x^5-1)/(x-1)=x^4+x^3+x^2+x+1 $. Thus this must divide$P $which implies$P(x)=4(x^4+x^3+x^2+x+1)\$. This completes the proof.

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