2012 AMC 12B Problems/Problem 23

Solution 1

Since $z_0$ 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: [1]). 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 $z=i$ . 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 $z=-i$ . 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 $z^2+z+1$ 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 $4x^4+4x^3+4x^2+4x$ $4x^4+4x^3+3x^2+3x$ $4x^4+4x^3+2x^2+2x$ $4x^4+4x^3+x^2+x$ $4x^4+4x^3$ $4x^4+4x^3+4x^2$ However, by inspection $4x^4+4x^3+4x^2+4x+4$ 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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2012 AMC 12B Problems/Problem 23

Solution 1