# Using least common multiple to prove there exists a prime between $2x$ and $3x$

Let $$\text{lcm}(x)$$ be the least common multiple of $$\{1,2,3,\dots, x\}$$.

Hanson showed that $$\text{lcm}(x) < 3^x$$

I'm wondering if the following argument is valid for showing that there is always a prime $$p$$ such that $$2x < p < 3x$$ for $$x \ge 246$$.

Here is my argument:

(1) if prime $$p$$ satisfies $$2x < p < 3x$$, then $$p | {{3x}\choose{2x}}$$.

This follows from the fact that $${{3x}\choose{2x}}$$ is an integer and that $$p$$ will not divide out by $$2x!$$ or $$x!$$.

(2) $${{3x}\choose{2x}} > \dfrac{6^x}{x}$$ for $$x \ge 4$$

For $$x \ge 4$$, $${{2x}\choose{x}} \ge \dfrac{4^x}{x}$$ since $${8\choose4} = 70 > \dfrac{4^4}{4} = 64$$ and $${{2x}\choose{x}} = 2\left(\dfrac{2x-1}{x}\right){2(x-1)\choose{x-1}} > 2\left(\dfrac{2x-1}{x}\right)\left(\dfrac{4^{x-1}}{x-1}\right) > \dfrac{4^x}{x}$$ and $${{3x}\choose{2x}} > \left(\dfrac{3x}{2x}\right)^x {{2x}\choose{x}} > \left(\dfrac{3}{2}\right)^x\dfrac{4^x}{x}$$

(3) For any prime $$q$$ such that $$\frac{3x}{2} \le q \le 2x$$, it follows that $$2q > 3x$$ and $$q \nmid {{3x}\choose{2x}}$$ since it will be divided out by $$2x!$$.

(4) Assume that there is no prime greater than $$2x$$ that divides $${{3x}\choose{2x}}$$

(5) It follows that $${{3x}\choose{2x}} < \text{lcm}(\frac{3x}{2})\text{lcm}(\sqrt{3x})$$ since:

• From Legendre's Formula, it is well known that if $$v_p(x)$$ is the highest power of $$p$$ that divides $$x$$, then $$v_p({{3x}\choose{2x}}) = \sum\limits_{i \le \log_p(3n)} \left\lfloor\frac{3x}{p^i}\right\rfloor - \left\lfloor\frac{2x}{p^i}\right\rfloor - \left\lfloor\frac{x}{p^i}\right\rfloor$$ which equals $$1$$ or $$0$$ for each $$i$$ so that $$v_p({{3x}\choose{2x}}) \le \log_p(3x)$$

• If a prime $$p > \sqrt{3x}$$, then $$v_p({{3x}\choose{2x}}) = 1$$ and $$p | \text{lcm}(\frac{3x}{2})$$

• If a prime $$p \le \sqrt{3x}$$, then $$v_p({{3x}\choose{2x}}) \le \log_p(3x) \le v_p(\frac{3x}{2})+1$$

(6) So that: $$\dfrac{6^x}{x} < 3^{3x/2}3^{\sqrt{3x}}$$

(7) But this is not true for $$x \ge 246$$ since:

$$246\ln 6 - \ln 246 > 435.26 > 435.24 > \frac{3}{2}(246)\ln 3 + \sqrt{3\times246}\ln 3$$

and for $$x \ge 246$$, $$6 > (3^{3/2})(3^{\sqrt{3x+2} - \sqrt{3x}})\left(\dfrac{x+1}{x}\right)$$

(8) So we can reject step (4).

• I'm sorry, but I made a mistake in my answer re: the last part of your statement (5). However, I've corrected it now so please reread that part and let me know if there are any other errors in it. Thanks. – John Omielan May 20 '19 at 21:26

Your proof provides an interesting use of Hanson's lcm inequality result and appears to be basically correct, but there are few small mistakes & other issues.

In your step (5), you wrote

... $$v_p({{3x}\choose{2x}}) = \sum\limits_{i \le \log_p(3n)} \left\lfloor\frac{3x}{p^i}\right\rfloor - \left\lfloor\frac{2x}{p^i}\right\rfloor - \left\lfloor\frac{x}{p^i}\right\rfloor$$ which equals $$1$$ or $$0$$ for each $$i$$ ...

First, in the summation part of $$i \le \log_p(3n)$$, the $$n$$ I believe is supposed to be $$x$$. Also, in something like a math journal, the statement "equals $$1$$ or $$0$$ for each $$i$$" may be obvious & sufficient, but I don't believe it is for here since, I at least, didn't initially know if it was always true. FYI, here is what I did to confirm this. I let $$d = p^i$$ and $$x = md + r$$ for $$m,d \in \mathbb{N}^0$$ and $$0 \le r \lt d$$. Then,

\begin{align} \left\lfloor \frac{3x}{d} \right\rfloor - \left\lfloor \frac{2x}{d} \right\rfloor - \left\lfloor \frac{x}{d} \right\rfloor & = 3m + \left\lfloor \frac{3r}{d} \right\rfloor - 2m - \left\lfloor \frac{2r}{d} \right\rfloor - m - \left\lfloor \frac{r}{d} \right\rfloor \\ & = \left\lfloor \frac{3r}{d} \right\rfloor - \left\lfloor \frac{2r}{d} \right\rfloor - \left\lfloor \frac{r}{d} \right\rfloor \\ & = \left\lfloor \frac{3r}{d} \right\rfloor - \left\lfloor \frac{2r}{d} \right\rfloor\tag{1}\label{eq1} \end{align}

Since $$r$$ and $$d$$ are positive, both terms are non-negative and $$\left\lfloor \frac{3r}{d} \right\rfloor \ge \left\lfloor \frac{2r}{d} \right\rfloor$$, so \eqref{eq1} is non-negative. Since $$\left\lfloor \frac{3r}{d} \right\rfloor \le 2$$, the only way \eqref{eq1} can be anything other than $$0$$ or $$1$$ would be for $$\left\lfloor \frac{3r}{d} \right\rfloor = 2$$ (which requires $$\frac{2d}{3} \le r \lt 1$$) and $$\left\lfloor \frac{2r}{d} \right\rfloor = 0$$ (which requires $$0 \le r \lt \frac{d}{2}$$). Since there is no overlap between those $$2$$ regions, the result of \eqref{eq1} must always be $$0$$ or $$1$$.

Next, you state

If a prime $$p > \sqrt{3x}$$, then $$v_p({{3x}\choose{2x}}) = 1$$ and $$p | \text{lcm}(\frac{3x}{2})$$

For $$x = 5$$, note that $$p = 5 \gt \sqrt{15}$$, but $${{3x}\choose{2x}} = {{15}\choose{10}} = 3 \times 7 \times 11 \times 13$$, so $$v_p({{3x}\choose{2x}}) = 0$$ in this case. A correct statement to make would be something like $$v_p({{3x}\choose{2x}}) \leq 1$$. Note this doesn't really affect your argument and it actually strengthens it slightly as you want to have a sufficiently small upper bound.

With the next statement of

If a prime $$p \le \sqrt{3x}$$, then $$v_p({{3x}\choose{2x}}) \le \log_p(3x) \le v_p(\frac{3x}{2})+1$$

first note I believe you mean the last part to be $$v_p\left(\text{lcm}\left(\frac{3x}{2}\right)\right) + 1$$ instead. However, for $$p = 2$$ and $$x = 4$$, then $$\log_p(3x) = \log_2(12) \gt 3$$, but $$v_p\left(\text{lcm}\left(\frac{3x}{2}\right)\right) + 1 = v_2(60) + 1 = 2 + 1 = 3$$. A correct statement could use the fact that $$v_p$$ is always an integer, so you can just use the integer component of the log to get $$v_p({{3x}\choose{2x}}) \le \left\lfloor \log_p(3x) \right\rfloor \le v_p\left(\text{lcm}\left(\frac{3x}{2}\right)\right) + 1$$.

Finally, for the second part of (7), you wrote

and for $$x \ge 246$$, $$6 > (3^{3/2})(3^{\sqrt{3x+2} - \sqrt{3x}})\left(\dfrac{x+1}{x}\right)$$

I don't see how you get this. Your step (6) states that $$\dfrac{6^x}{x} < 3^{3x/2}3^{\sqrt{3x}}$$. If you multiply by $$x$$ and take the $$x$$'th root of both sides, you get $$6 \lt 3^{3/2}3^{\sqrt{3/x}}x^{1/x}$$. I suggest not only showing how you got your result, but also how you proved it to be always be true for $$x \ge 246$$.

Here is how I would prove that (6) is not true for $$x \ge 246$$. First, note that since logarithms, including the natural logarithm, are a strictly increasing function for positive numbers, so $$a \ge b \iff \log a \ge \log b$$. Thus, take the natural logarithm of both side of (6) and move the right side to the left side to get the following function

\begin{align} f(x) & = \ln(6)x - \ln(x) - \frac{3\ln(3)}{2}x - \ln(3)\sqrt{3}\sqrt{x} \\ & = \left(\ln(6) - \frac{3\ln(3)}{2}\right)x - \ln(3)\sqrt{3}\sqrt{x} - \ln(x) \tag{2}\label{eq2} \end{align}

You've already shown that $$f(246) \gt 0$$. If can show that $$f'(x) \ge 0$$ for $$x \ge 246$$, then $$f(x) \gt 0$$ for all $$x \ge 246$$. Differentiating \eqref{eq2} gives

\begin{align} f'(x) & = \left(\ln(6) - \frac{3\ln(3)}{2}\right) - \left(\frac{\ln(3)\sqrt{3}}{2}\right) \frac{1}{\sqrt{x}} - \frac{1}{x} \\ & = \frac{1}{x}\left(\left(\ln(6) - \frac{3\ln(3)}{2}\right)x - \left(\frac{\ln(3)\sqrt{3}}{2}\right)\sqrt{x} - 1\right) \tag{3}\label{eq3} \end{align}

Note that

$$g(x) = \left(\ln(6) - \frac{3\ln(3)}{2}\right)x - \left(\frac{\ln(3)\sqrt{3}}{2}\right)\sqrt{x} - 1 \tag{4}\label{eq4}$$

is a quadratic equation in $$\sqrt{x}$$. As the coefficient of $$x$$, i.e., $$\ln(6) - \frac{3\ln(3)}{2} = 0.143841\ldots \gt 0$$, the value will always be positive for large enough $$x$$. Let $$y = \sqrt{x}$$ to transform \eqref{eq4} to

$$h(y) = \left(\ln(6) - \frac{3\ln(3)}{2}\right)y^2 - \left(\frac{\ln(3)\sqrt{3}}{2}\right)y - 1 \tag{5}\label{eq5}$$

Note that $$h(\sqrt{246}) = 19.462358\ldots$$. Also,

$$h'(y) = 2\left(\ln(6) - \frac{3\ln(3)}{2}\right)y - \left(\frac{\ln(3)\sqrt{3}}{2}\right) \tag{6}\label{eq6}$$

Since $$h'(\sqrt{246}) = 3.560690\ldots$$ and $$h'(y)$$ is a strictly increasing linear function, this shows that, in summary, $$f(x)$$ in \eqref{eq2} is always positive for $$x \ge 246$$, which confirms what you're trying to prove.