efang wrote:

Holddddd up if you're saying I spent 30 minutes on this freaking problem just to get it wrong at midnight I will quite possibly fall into depression and eat all of my halloween candy.

First off $x = -2$ doesn't even work. Nor does $x = \frac{-7}{4}, -\frac{3}{2}, 2, \frac{9}{4}, \frac{5}{2},$ and $\frac{11}{4}$ (trust me I calculated each of those individually to check for any mistakes).

Here is my solution:

The restriction of $\lfloor{x^3}\rfloor = 4x+3$ tells us that $4x+3 \le x^3 < 4x+4$ . It also tells us that $4x + 3$ must be an integer which means that $4x$ is an integer.

This means that $x = \frac{a}{4}$ for some integer $a$ .

First let's investigate $4x +3 \le x^3$

By rearranging the numbers we have $x^3 - 4x - 3 \ge 0$ . By RRT we find that $-1$ is a root of $x^3 - 4x - 3$ . Factoring we get $(x+1)(x^2-x-3) \ge 0$ .

By the quadratic equation the solutions for $x^2 - x - 3 =0$ are $x = \frac{1 \pm \sqrt{13}}{2}$ .

Thus this condition is satisfied if $\frac{1-\sqrt{13}}{2} \le x \le -1$ or $\frac{1+\sqrt{13}}{2} \le x$

Now let's investigate $x^3 < 4x + 4$

We have that when $x = 2, 8 < 12$ but when $x = 3, 27 > 16$ . Obviously $x^3$ increases faster than $4x + 4$ so $x < 3$ . Note that this bound is not strong and in fact $x \le c$ for some $2 < c < 3$

Now we want to find the values of $x$ in the range $\frac{1-\sqrt{13}}{2} \le x \le -1$ such that $x = \frac{a}{4}$ for some integer $a$ .

Note that $3.6^2 = 12.96 \approx 13$ . Thus $\frac{1-\sqrt{13}}{2} \approx \frac{1-3.6}{2} = -1.3$

Thus the possible values of $x$ in this range are $-1.25, -1$ which we can check to see that they both work.

Now we approximate $\frac{1+\sqrt{13}}{2} \approx \frac{1+3.6}{2} = 2.3$

Thus the possible values of this range are $2.5$ and $2.75$ . But wait! We have to check if these values indeed satisfy $x^3 - 4x - 4 < 0$ because we never determined the value $a$ such that $x \le a$ . Plugging in $2.5$ we see that the condition is not satisfied. As again, $x^3$ grows faster than $4x+4, x=2.75$ won't work either.

Thus the solutions are simply $\boxed{x = -1.25, -1}$

Hm it's too late right now (even with daylight savings) for me to have the energy to read the above solution so I might do that later and try finding what's wrong.

efang's right. Also, you could have noted that the fractional part of x lies in the set S = { $0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}$ } since 4x+3 must be an integer. From there you note that $n^3 < 4n+3 + 4s < (n+1)^3$ , where x = n+s, n is an integer, and s $\in$ S. Separating the bound into two polynomials and casework bashing gives you x = -1.25 and -1.