# Difference between revisions of "2012 AMC 12B Problems/Problem 23"

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− | == | + | == Problem 23 == |

− | + | Consider all polynomials of a complex variable, <math>P(z)=4z^4+az^3+bz^2+cz+d</math>, where <math>a,b,c,</math> and <math>d</math> are integers, <math>0\le d\le c\le b\le a\le 4</math>, and the polynomial has a zero <math>z_0</math> with <math>|z_0|=1.</math> What is the sum of all values <math>P(1)</math> over all the polynomials with these properties? | |

− | + | <math>\textbf{(A)}\ 84\qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 108\qquad\textbf{(E)}\ 120 </math> | |

− | + | == Solution == | |

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

− | Finally suppose <math>z^2+z+1< | + | Second, assume that </math>z_0\in \mathbb{C} \backslash \mathbb{R}<math>. |

+ | |||

+ | ==Solution (doubtful) == | ||

+ | |||

+ | Since </math>z_0<math> is a root of </math>P<math>, and </math>P<math> has integer coefficients, </math>z_0<math> must be algebraic. Since </math>z_0<math> is algebraic and lies on the unit circle, </math>z_0<math> 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 </math>P<math> has degree 4, it seems reasonable (and we will assume this only temporarily) that </math>z_0<math> must be a 2nd, 3rd, or 4th root of unity. These are among the set </math>\{\pm1,\pm i,(-1\pm i\sqrt{3})/2\}<math>. Since complex roots of polynomials come in conjugate pairs, we have that </math>P<math> has one (or more) of the following factors: </math>z+1<math>, </math>z-1<math>, </math>z^2+1<math>, or </math>z^2+z+1<math>. If </math>z=1<math> then </math>a+b+c+d+4=0<math>; a contradiction since </math>a,b,c,d<math> are non-negative. On the other hand, suppose </math>z=-1<math>. Then </math>(a+c)-(b+d)=4<math>. This implies </math>a+b=8,7,6,5,4<math> while </math>b+d=4,3,2,1,0<math> correspondingly. After listing cases, the only such valid </math>a,b,c,d<math> are </math>4,4,4,0<math>, </math>4,3,3,0<math>, </math>4,2,2,0<math>, </math>4,1,1,0<math>, and </math>4,0,0,0<math>. | ||

+ | |||

+ | Now suppose </math>z=i<math>. Then </math>4=(a-c)i+(b-d)<math> whereupon </math>a=c<math> and </math>b-d=4<math>. But then </math>a=b=c<math> and </math>d=a-4<math>. This gives only the cases </math>a,b,c,d<math> equals </math>4,4,4,0<math>, which we have already counted in a previous case. | ||

+ | |||

+ | Suppose </math>z=-i<math>. Then </math>4=i(c-a)+(b-d)<math> so that </math>a=c<math> and </math>b=4+d<math>. This only gives rise to </math>a,b,c,d<math> equal </math>4,4,4,0<math> which we have previously counted. | ||

+ | |||

+ | Finally suppose </math>z^2+z+1<math> divides </math>P<math>. Using polynomial division ((or that </math>z^3=1<math> to make the same deductions) we ultimately obtain that </math>b=4+c<math>. This can only happen if </math>a,b,c,d<math> is </math>4,4,0,0<math>. | ||

Hence we've the polynomials | Hence we've the polynomials | ||

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<cmath>4x^4+4x^3</cmath> | <cmath>4x^4+4x^3</cmath> | ||

<cmath>4x^4+4x^3+4x^2</cmath> | <cmath>4x^4+4x^3+4x^2</cmath> | ||

− | However, by inspection <math>4x^4+4x^3+4x^2+4x+4< | + | However, by inspection </math>4x^4+4x^3+4x^2+4x+4<math> has roots on the unit circle, because </math>x^4+x^3+x^2+x+1=(x^5-1)/(x-1)<math> 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 </math>z_0<math> 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 </math>z_0<math> in an </math>n<math>th root of unity where </math>n>5<math>, and </math>z_0<math> is not a 3rd or 4th root of unity. (Note that 1st and 2nd roots of unity are themselves 4th roots of unity). If </math>n<math> is prime, then \textit{every} </math>n<math>th root of unity except 1 must satisfy our polynomial, but since </math>n>5<math> and the degree of our polynomial is 4, this is impossible. Suppose </math>n<math> is composite. If it has a prime factor </math>p<math> greater than 5 then again every </math>p<math>th root of unity must satisfy our polynomial and we arrive at the same contradiction. Therefore suppose </math>n<math> is divisible only by 2,3,or 5. Since by hypothesis </math>z_0<math> is not a 2nd or 3rd root of unity, </math>z_0<math> 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 </math>P(z_0)=0<math>. But </math>(x^5-1)/(x-1)<math> has exactly all 5th roots of unity excluding 1, and </math>(x^5-1)/(x-1)=x^4+x^3+x^2+x+1<math>. Thus this must divide </math>P<math> which implies </math>P(x)=4(x^4+x^3+x^2+x+1)$. This completes the proof. |

## Revision as of 02:31, 6 December 2012

## Problem 23

Consider all polynomials of a complex variable, , where and are integers, , and the polynomial has a zero with What is the sum of all values over all the polynomials with these properties?

## Solution

First, assume that , so or . does not work because . Assume that . Then , we have , so a=44+b=a+c \leq a+b4+b=4+cb=c(a,b,c,d)(4,t,t,0)P(-1)=4-4+t-t+0=0P(1)\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_0PPz_0z_0z_0Pz_0\{\pm1,\pm i,(-1\pm i\sqrt{3})/2\}Pz+1z-1z^2+1z^2+z+1z=1a+b+c+d+4=0a,b,c,dz=-1(a+c)-(b+d)=4a+b=8,7,6,5,4b+d=4,3,2,1,0a,b,c,d4,4,4,04,3,3,04,2,2,04,1,1,04,0,0,0$.

Now suppose$ (Error compiling LaTeX. ! Missing $ inserted.)z=i4=(a-c)i+(b-d)a=cb-d=4a=b=cd=a-4a,b,c,d4,4,4,0$, which we have already counted in a previous case.

Suppose$ (Error compiling LaTeX. ! Missing $ inserted.)z=-i4=i(c-a)+(b-d)a=cb=4+da,b,c,d4,4,4,0$which we have previously counted.

Finally suppose$ (Error compiling LaTeX. ! Missing $ inserted.)z^2+z+1Pz^3=1b=4+ca,b,c,d4,4,0,0$.

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+4x^4+x^3+x^2+x+1=(x^5-1)/(x-1)z_0z_0nn>5z_0nnn>5nppnz_0z_0P(z_0)=0(x^5-1)/(x-1)(x^5-1)/(x-1)=x^4+x^3+x^2+x+1PP(x)=4(x^4+x^3+x^2+x+1)$. This completes the proof.