Let $\lfloor x \rfloor = n, \{ x \} = f$ then: $n^3 - 7 \left \lfloor \left( n+f + \frac13 \right) \right \rfloor = -13$ We make cases: Case 1: $f < \frac23 \implies n^3 - 7n + 13 = 0$ no solutions Case 2: $f \geq \frac23 \implies n^3 - 7(n+1) = - 13$ $\implies n^3 - 7n + 6 = 0 \implies n = 1, 2, -3$ Note that there is no additional restriction on $f$. So solutions are: $\boxed{\{ n + f : n = \{1,2,-3\} \text{ and } \frac23 \leq f < 1 \}} $

I am very confused. Why can't the interval for f be from [2/3,5/3)?

club52 wrote: I am very confused. Why can't the interval for f be from [2/3,5/3)? Since the solution defined f as the fractional part of x, and by definition, the bounds of f are [0, 1)

Hexagrammum16 wrote: Let $\lfloor x \rfloor = n, \{ x \} = f$ then: $n^3 - 7 \left \lfloor \left( n+f + \frac13 \right) \right \rfloor = -13$ We make cases: Case 1: $f < \frac23 \implies n^3 - 7n + 13 = 0$ no solutions Case 2: $f \geq \frac23 \implies n^3 - 7(n+1) = - 13$ $\implies n^3 - 7n + 6 = 0 \implies n = 1, 2, -3$ Note that there is no additional restriction on $f$. So solutions are: $\boxed{\{ n + f : n = \{1,2,-3\} \text{ and } \frac23 \leq f < 1 \}} $ Why "f" use to here? I don't understand.

mathsolver01 wrote: Hexagrammum16 wrote: Let $\lfloor x \rfloor = n, \{ x \} = f$ then: $n^3 - 7 \left \lfloor \left( n+f + \frac13 \right) \right \rfloor = -13$ We make cases: Case 1: $f < \frac23 \implies n^3 - 7n + 13 = 0$ no solutions Case 2: $f \geq \frac23 \implies n^3 - 7(n+1) = - 13$ $\implies n^3 - 7n + 6 = 0 \implies n = 1, 2, -3$ Note that there is no additional restriction on $f$. So solutions are: $\boxed{\{ n + f : n = \{1,2,-3\} \text{ and } \frac23 \leq f < 1 \}} $ Why "f" use to here? I don't understand. We don't normally use "f" for $\left\{x \right\}$, but that won't matter. Still, I recommend using $\alpha$.

Let $\lfloor x \rfloor = n, \left\{x\right\} = \alpha \left(0\leq\alpha<1\right)$ Case 1: $\lfloor x+\frac13 \rfloor = n $ It means $ 0 \leq \alpha < \frac23 , n^3 - 7n = -13 $ and it does not have any interger solutions. Case 2: $\lfloor x+\frac13 \rfloor = n + 1$ It means $ \frac23 \leq \alpha < 1, n^3 - 7\left(n+1\right) = -13.$ $n^3 - 7\left(n+1\right) + 13 = n^3 - 7n +6 = 0 $ has interger solutions of $ n = -3, 1, 2$ Thus, this equation holds for x that satisfies $\lfloor x \rfloor = -3, 1, 2 $ and $\frac23\leq\left\{x\right\}<1 $ $\boxed{\therefore -\frac73 \leq x < -2 \text{ or } \frac53 \leq x < 2 \text{ or } \frac83 \leq x < 3}$