Greatest Integer Function and graphing

Question

Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ and let $n^3 + 3n = 6 (\lfloor n \rfloor ^2) + \frac {478}{27}$, then solve for $n$? Using plot function in wolfarm gives me different solution when I try to solve analytically. Also when I try to impose condition that fractional part is less than 1 but greater than equal to zero, the equation turns out to be unwieldy for me. What is the right way to tackle this problem?

Answer 1

Let $$f(x) = x^3 - 6 \lfloor x \rfloor^2 + 3x - 478/27 \tag{A}$$

Then we are trying to solve $f(x) = 0$. Suppose that $x = n + \delta$, where $n \in \mathbb Z$ and $\delta \in [0,1)$.

Note that, for $n \le x < n+1$, $\frac{d}{dx} \lfloor x \rfloor = 0$.

1. For $x \in [n, n+1)$, $f'(x) = 3x^2+3 > 0$.
2. On each interval, $x \in [n, n+1)$, f(x) is increasing.
3. $f(6) = \dfrac{8}{27}$.
4. For all $x \ge 6$, $f(x) > 0$.
5. $f(3) = -\frac{964}{27}$ and $f(3.99)=3.7875$
6. For all $x \le 3$, $f(x) < 0$.

It follows that $f(x) = 0$ implies that $3 < x < 6$.

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$x=3+\delta$

$\qquad f(x) = \delta^3 + 9\delta^2 + 30\delta - 964/27$

$\qquad f(x) = 0 \implies x \approx 3.91403$

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$x=4+\delta$

$\qquad f(x)= \delta^3 + 12\delta^2 + 51\delta - \dfrac{101}{27}$

$\qquad f(x) = 0 \implies x \approx 4.072117 $

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$x=5+\delta$

$\qquad f(x) = \delta^3 + 15\delta^2 + 78\delta - 748/27$

$\qquad f(x) = 0 \implies x = \dfrac{16}{3}$