Countably many

Assume that $R \ge 0$ , so that $\sqrt{R}$ is real. Write $L=5\times10^8$ , and suppose that $K = (L+\sqrt{R})^{\frac13} + (L-\sqrt{R})^{\frac13}$ is an integer. It is clear that $K \ge 0$ . Then $\begin{array}{rcl} K^3 & = &L + \sqrt{R} + 3K(L^2 - R)^{\frac13} + L - \sqrt{R} \\ & = & 2L + 3K(L^2 - R)^{\frac13} \end{array}$ and hence $(K^3 - 2L)^3 \; = \; 27K^3(L^2 - R)$ It is clear from this equation that $K \neq 0$ , so $K \ge 1$ . Then the requirement that $R \ge 0$ tells us that $\begin{array}{rcl} (K^3 - 2L)^3 & \le & 27K^3L^2 \\ K^3 - 2L & \le & 3L^{\frac23}K \\ K^3 - 3L^{\frac23}K - 2L & \le & 0 \\ \big(K + L^{\frac13}\big)^2\big(K - 2L^{\frac13}\big) & \le & 0 \end{array}$ so we deduce that $1 \le K \le 2L^{\frac13}$ .

Now $R \; = \; L^2 - \left(\frac{K^3 - 2L}{3K}\right)^3$ and hence $\frac{dR}{dK} \; = \; -3\left(\frac{K^3 - 2L}{3K}\right)^2\left(\tfrac23K + \frac{2L}{3K^2}\right) \; < \; 0$ for all $K > 0$ . Thus we obtain a different rational value of $R$ for every different positive integer value of $K$ . Thus we deduce that there are $\big\lfloor 2L^{\frac13} \big\rfloor$ possible different values of $R$ . For our value of $L$ , this means that there are $1587$ possible different values of $R$ , making the desired answer $587$ .

We rely on our traditional method of finding roots of surds.

Let $(5 \times 10^8) + \sqrt{R} = (a+\sqrt{b})^3$ . Expanding RHS, & equating rational & irrational coefficients, we get:

$a^3 + 3ab=(5 \times 10^8) \Rightarrow b= \frac{(5 \times 10^8)-a^3}{3a}$ .

Also, $\sqrt{R}= (3a^2+b)\sqrt{b} \Rightarrow R = (3a^2+b)^2 (b) = \frac{((5 \times 10^8)+8a^3)^2((5 \times 10^8)-a^3)}{27a^3}$ . [on substituting for $b$ ]

Clearly $a$ must be positive, for $R$ to be positive. Also, $a^3 < (5 \times 10^8) \Rightarrow a < \sqrt[3]{.5} \times 10^3 < 793.701$ . For each value of $a$ , we get a rational $R$ .

The expression in the question evaluates to $2a$ . Thus for $2a$ to be integral, either $a$ is integral, or is rational with denominator $2$ .

• Case 1: $a \in \mathbb{N}$ . Then $a$ can be any of $1,2,...,793$ giving $793$ choices.

• Case 2: $a= \frac{p}{2}$ (where $p$ is odd) Then $p < 2 \times 793.701 = 1587.402 \Rightarrow 1 \leq p \leq 1587$ . There are $794$ odd numbers from $1$ to $1587$ (inclusive) & $p$ can be any of them.

Thus, there can be $793+794 = 1587$ values of $a$ ,& correspondingly that many values for $R$ .