2007 iTest Problems/Problem 46

Problem

Let $(x,y,z)$ be an ordered triplet of real numbers that satisfies the following system of equations: $\begin{align*}x+y^2+z^4&=0,\\y+z^2+x^4&=0,\\z+x^2+y^4&=0.\end{align*}$ If $m$ is the minimum possible value of $\lfloor x^3+y^3+z^3\rfloor$ , find the modulo $2007$ residue of $m$ .

Solution

Rearrange the terms to get $y^2 + z^4 = -x$ $z^2 + x^4 = -y$ $x^2 + y^4 = -z$ Since the left hand side of all three equations is greater than or equal to 0, $x,y,z \le 0$ . Also, note that the equations have symmetry, so WLOG, let $0 \ge x \ge y \ge z$ . By substitution, we have

$y^2 + z^4 \le z^2 + x^4 \le x^2 + y^4$

Note that $0 \le x^2 \le y^2 \le z^2$ and $0 \le x^4 \le y^4 \le z^4$ . That means $x^2 + y^4 \le x^2 + z^4$ . Since $y^2 + z^4 \le x^2 + y^4$ , $y^2 + z^4 \le x^2 + z^4$ $y^2 \le x^2$ Since $x^2 \le y^2$ , then $x^2 = y^2$ . Because $x$ and $y$ are nonpositive, $x = y$ .

Using substitution in the original system, $x^2 + z^4 = z^2 + x^4$ $(z^2 + x^2)(z^2 - x^2) - (z^2 - x^2) = 0$ $(z^2 - x^2)(x^2 + z^2 - 1) = 0$ To find the real solutions, we use casework and the Zero Product Property.

Case 1: $z^2 = x^2$

If $z^2 = x^2$ , then since $z$ and $x$ are nonpositive, then $z = x$ . Substitution results in $x+x^2+x^4 = 0$ $x(1+x+x^3) = 0$ That means $x = 0$ or $x^3 + x + 1 = 0$ . For the first equation, $m = 0$ . For the second equation, note that $x^3 = -x-1$ , and since $x = y = z$ , $m = \lfloor -3x-3 \rfloor$ , where $x$ is a real number. Since $-\tfrac{1}{3}^3 - \tfrac13 + 1 = \tfrac{16}{27}$ and $-\tfrac{2}{3}^3 - \tfrac23 + 1 = \tfrac{1}{27}$ , the root of $x$ is less than $-\tfrac23$ but more than $-1$ , so $0 > -3x-3 > -1$ $m = \lfloor -3x-3 \rfloor = -1$

Case 2: $x^2 + z^2 = 1$

Because $x^2 \le z^2$ , $x^2 \le \frac12$ . From one of the original equations, $z^2 + x^4 = -y$ $1 - x^2 + x^4 + x = 0$ Using the Rational Root Theorem, $(x+1)(x^3 - x^2 + 1) = 0$ Note that if $x = -1$ , then $x^2 \ge \tfrac12$ , so that won’t work. Let $x = -a$ (where $a \ge 0$ since $x \le 0$ ), so $a^3 + a^2 = 1$ If $a \le \tfrac{\sqrt2}{2}$ , then $a^3 + a^2 \le \frac{\sqrt2}{4} + \frac12$ $a^3 + a^2 \le \frac{\sqrt2 + 2}{4} < 1$ Thus, there are no solutions in this case.

From the two cases, the smallest possible value of $m$ is $-1$ , so the modulo $2007$ residue of $m$ is $\boxed{2006}$ .

2007 iTest (Problems, Answer Key)
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2007 iTest Problems/Problem 46

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