, 2010

Algebra Difficulty 5.4 AIME, harder Prove it Ukraine

Let $P(x)$ , $Q(x)$ and $R(x)$ be polynomials, such that $Q(x)$ and $R(x)$ take nonnegative values only. It is known, that the equation
$P(x) + \sqrt{Q(x)} + \sqrt{Q(x) + \sqrt{R(x)}} = 0$
has infinitely many solutions. Is it true, that every real number is a root of this equation?

Solution

Not necessary.

We construct the following example: $P(x) = x$ , $Q(x) = \frac{1}{4}x^2$ , $\sqrt{Q(x)} = \frac{1}{2}|x|$ , $R(x) = 0$ . Then, $\sqrt{Q(x) + \sqrt{R(x)}} = \frac{1}{2}|x|$ and we have:
$P(x) + \sqrt{Q(x)} + \sqrt{Q(x) + \sqrt{R(x)}} = x + |x| = \begin{cases} 2x, & x \ge 0, \\ 0, & x < 0. \end{cases}$
Hence, our left hand side does not equal to $0$ for all $x$ , but the equation has infinitely many solutions.

Note, that this example is not unique, one can also take $Q(x) = \frac{1}{8}x^2$ , $R(x) = \left(\frac{1}{8}x^2\right)^2$ .

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.