Sketch: Clearly $ x$ is a rational number, moreover, $ x$ is of the form $ \frac{k-3}{4}$. Now after considering $ 4$ cases of what residue $ k$ gives modulo $ 4$, we're about to get $ 4$ equations.

This problem seems unanswered yet, so i will write my solution. Corrections are always welcome. Thanks Notice that the problem is equivalent to \[x = \frac{\lfloor x^3 \rfloor - 3}{4} \longleftrightarrow \lfloor x \rfloor = \lfloor \frac{\lfloor x^3 \rfloor - 3}{4} \rfloor \] Also, notice that \[(x+1)(x^2-x-3) = x^3-4x-3 \geq \lfloor x^3 \rfloor -4x-3 = 0\] \[(x+1)(x^2-x-3) \geq 0\] This gives \[\frac{1-\sqrt{13}}{2} \leq x \leq -1 \text{ or } x \geq \frac{1+\sqrt{13}}{2}\] So, \[-2 \leq \lfloor x \rfloor \leq -1 \text{ or} \lfloor x \rfloor \geq 2\] Now, consider the function $f(x) = x^3 - 4x - 3$. We will find on which interval does $f$ increase, so we need to take derivative, then $f'(x) > 0 \longleftrightarrow 3x^2 - 4 > 0 \longleftrightarrow (x\sqrt{3} - 2)(x\sqrt{3} + 2) > 0$, then $-\frac{2}{3}\sqrt{3} \geq x \text{ or } x \geq \frac{2}{3}\sqrt{3}$. Notice again that $f(3) = 27 - 12 - 3 = 12 > 0$. Since $3 > \frac{2}{3}\sqrt{3}$, and $f$ is increasing on $[\frac{2}{3}\sqrt{3}, +\infty)$, then $f(x) > 0$, so $x < 3$. Noticing that $f(-2) = -8 + 8 - 3 = -3 < 0$ and using similar reasoning, gives $x > -2$. So, we have \[-2 \leq \lfloor x \rfloor \leq -1 \text{ or } 2 \leq \lfloor x \rfloor < 3\] This means, the possible values for $\lfloor x \rfloor$ are $-2 , -1, 2$ If $\lfloor x \rfloor = -2$, then $-2 \leq x < -1$, so $-8 \leq x^3 < -1$ and $-8 \leq \lfloor x^3 \rfloor < -1$. So, $\lfloor x^3 \rfloor = -8, -7, -6, ..., -2$. Plugging each value to $\lfloor x \rfloor = \lfloor \frac{\lfloor x^3 \rfloor - 3}{4} \rfloor$, we see that the only possible values are $-5, -4, -3, -2$. If $\lfloor x \rfloor = -1$, then $-1 \leq x < 0$, so $-1 \geq x^3 < 0$ and $1 \leq \lfloor x^3 \rfloor < 0$. Checking, we see that $-1$ is the only possible value for $\lfloor x^3 \rfloor$ in this case. If $\lfloor x \rfloor = 2$, then $2 \leq x \leq 3$, so $8 \geq x^3 < 27$ and $8 \leq \lfloor x^3 \rfloor < 27$. Thus, $\lfloor x^3 \rfloor = 8, 9, 10, \cdots, 26$. Checking again, we see that the possible values are $11,12,13,14$. So \[\lfloor x^3 \rfloor = -5, -4, -3, -2, -1, 11, 12, 13, 14\] Hence, since $x = \frac{\lfloor x^3 \rfloor - 3}{4}$, then we substitute those values there and get : \[\boxed{x = -2, -\frac{7}{4}, -\frac{3}{2}, -\frac{5}{4}, -1, 2, \frac{9}{4}, \frac{5}{2}, \frac{11}{4}}\]

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.

I don't think there's anything wrong in MrRTI's solution. He just gave a correct proof for reducing the set of possible solutions to the 9 numbers he stated as "solutions". The mistake is just that not all these numbers really satisfy the equation because the conditions you elaborated are just necessary but not sufficient ones. So, you just need to plug in all your 9 values to see that only $1.25$ and $1$ will be solutions.

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.

I think I have a somehow easier to understand (for people more in the beginners level like me) solution that isn't too long

Brb. Ok done

orl wrote: Find all real numbers $ x$ such that $ \lfloor x^3 \rfloor = 4x + 3$. Problem is "find all $n\in\mathbb Z$ such that $n+4>\frac {n^3}{64}\ge n+3$" (and then $x=\frac n4$) Which is $256>n^3-64n\ge 192$ And it is not very difficult to get that only $n\in\{-5,-4\}$ fits, and so $x\in\{-\frac 54,-1\}$

This is what I got ( I think so it is wrong) Observe LHS is an integer so rhs should also be. so is 4x so is of always of the form k/4