prove that $(p,mp]$ will always contain at least $y-1$ cubes

If $n, m\in \mathbb{Z}_{>0}$ and $(n, mn]$ contains $y$ cubes, prove that $(p,mp]$ will always contain at least $y-1$ cubes (integers $a$ so that $a=k^3$ for some integer k) for all $p\in [n,\infty)$.

This works for small examples, such as for $n=7, m=4$ when the cubes are $8$ and $27$ as $(p, 4p]$ always contains at least one cube for $p\ge n$, but I'm not sure how to generalize the results. I know that between any two real numbers that are at least one apart there is an integer. Also, clearly, the ratio between $k^3$ and $(k+1)^3$ decreases as $k$ increases. For $n\ge 8$, there exists an integer $k$ so that $k \in (n, mn]$ is a perfect cube because $\sqrt[3]{mn} - \sqrt[3]{n} \ge \sqrt[3]{n} > 1$.

Solution 1:

Lemma (without proof): Show that $ ( n, mn ] $ contains at least $y$ cubes iff $$ \lfloor \sqrt[3]{mn} \rfloor - \lfloor \sqrt[3] {n} \rfloor \geq y.$$

Corollary: The problem can be restated to:

For $ p \geq n \geq 1$, and $ m \geq 1$, show that
$$ \lfloor \sqrt[3]{mp} \rfloor - \lfloor \sqrt[3] {p} \rfloor \geq \lfloor \sqrt[3]{mn} \rfloor - \lfloor \sqrt[3] {n} \rfloor - 1.$$

Proof of inequality: This is true because with $ \sqrt[3]{m} \geq 1$,

$$ \lfloor \sqrt[3]{mp} \rfloor - \lfloor \sqrt[3]{mn} \rfloor \geq \lfloor \sqrt[3]{mp} - \sqrt[3]{mn} \rfloor \geq \lfloor \sqrt[3]{m} \times \lfloor \sqrt[3]{p} - \sqrt[3]{n} \rfloor \rfloor \\ \geq \lfloor \sqrt[3]{p} - \sqrt[3]{n} \rfloor \geq \lfloor \sqrt[3]{p} \rfloor - \lfloor\sqrt[3]{n} \rfloor - 1.$$

The absolute-function inequalities used at each stage of the inequality are derived from the well-known inequalities (for suitable values of $x, y$, and rearranged as needed):

• $ \lfloor x \rfloor + \lfloor y \rfloor \leq \lfloor x+y \rfloor$
• $ \lfloor xy \rfloor \geq \lfloor x \lfloor y \rfloor \rfloor$.
• For $ x \geq 1$, $ \lfloor x \lfloor y \rfloor \rfloor \geq \lfloor y \rfloor$
• $ \lfloor x \rfloor + \lfloor y \rfloor \geq \lfloor x+y \rfloor - 1$

Solution 2:

The statement to prove is obviously true for $y = 0$ and $y = 1$, so consider just $y \gt 1 \; \to \; y - 1 \gt 0$. Since $(n, mn]$ contains $y$ cubes, this means there's an $a \in \mathbb{Z}_{>0}$ such that

$$n \lt a^3 \lt (a + (y - 1))^3 \le mn \tag{1}\label{eq1A}$$

Since the ratio of the upper & lower limits, i.e., $mn$ and $n$, must be greater than the ratio of the two intermediate values, this leads to

$$\frac{mn}{n} = m \gt \left(\frac{a + y - 1}{a}\right)^3 = \left(1 + \frac{y-1}{a}\right)^3 \tag{2}\label{eq2A}$$

For $n \le p \lt a^3$, no cubes will become out of range of $(p, mp]$ at the bottom or top, so there will be at least $y$ cubes in $(p, mp]$. Next, for $p \in [a^3, \infty)$, let $b$ be the number of cubes in $(p, mp]$. Thus, for some integer $c \ge 1$, we have $p \lt (a + c)^3$ and $(a + c + (b - 1))^3 \le mp$, plus also

$$(a + c - 1)^3 \le p \lt mp \lt (a + c + b)^3 \tag{3}\label{eq3A}$$

Handling the ratios as discussed for \eqref{eq1A} gives that

$$\left(\frac{a + c + b}{a + c - 1}\right)^3 = \left(1 + \frac{b + 1}{a + c - 1}\right)^3 \gt \frac{mp}{p} = m \tag{4}\label{eq4A}$$

Since $f(x) = x^3$ is a strictly increasing function for $x \gt 0$, from \eqref{eq2A} and \eqref{eq4A} we have

$$\left(1 + \frac{b + 1}{a + c - 1}\right)^3 \gt m \gt \left(1 + \frac{y-1}{a}\right)^3 \; \to \; \frac{b + 1}{a + c - 1} \gt \frac{y - 1}{a} \tag{5}\label{eq5A}$$

Since $c \ge 1$, then $a + c - 1 \ge a$, so

$$b + 1 \gt y - 1 \; \to \; b \ge y - 1 \tag{6}\label{eq6A}$$

This concludes proving that for all $p \in [n, \infty)$, there are at least $y - 1$ cubes in $(p, mp]$.

Note the above proof techniques apply not only to cubes, they also work for any positive power (e.g., "cubes" may be replaced by "squares").