# 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.