# Prove that there is no positive integer $C$ and the sequence of positive integers$(a_n)$ satisfies : ${a_{k+1}}^k|C^k(a_1.a_2...a_k)$ for all $k$

Prove that there is no positive integer $$C$$ and the sequence of positive integers$$(a_n)$$ satisfies : $${a_{k+1}}^k|C^k(a_1.a_2...a_k)$$ for all $$k\in Z+$$ ($$(a_n)$$ is the sequence of numbers such that all numbers are DIFFERENT.)

Suppose there exists a positive integer $$C$$ and such a sequence $$(a_n)$$ .

Consider any prime $$p$$. Set $$vp(C) = C' , vp(a_i)=b_i$$

From the assumption $$\Rightarrow kb_{k+1} \le kC'+b_1+b_2+...+b_k$$

Consider the sequence $$(v_n)$$ as follows: $$v_1=b_1 ; kv_{k+1} = kC'+v_1+v_2+...+v_k$$ Thus, $$b_k \le v_k$$

$$kv_{k+1}=kC'+v_1+v_2+...+v_k=C'+(k-1)C'+v_1+v_2+...+v_{k-1}+v_k = C'+ (k-1)v_k+v_k=C'+kv_k$$ $$\Rightarrow v_{k+1} = C'/k + v_k$$

So if $$k→+°$$ , then because $$b_k \in Z+$$ $$(b_n)$$ is bounded on .

So $$(b_n)$$ is bounded above and below .

There should be an infinite $$i$$ such that there exists an infinite $$j$$ such that $$b_i = b_j$$

This is all I can do , this is a very difficult problem for me . Hope to get help from everyone. Thanks very much !

• What if $C = 1$ and $a_k = 1$ for all positive integers $k$? Feb 19, 2022 at 6:52
• @VTand sorry so much! (a_n) is the sequence of numbers such that all numbers are DIFFERENT. Thank you ! Feb 19, 2022 at 6:56

## 1 Answer

You're asking to prove there's no positive integer $$C$$ and an infinite set of distinct positive integers $$a_i$$ such that

$$(a_{k+1})^k \mid C^k\left(\prod_{i=1}^{k}a_i\right) \tag{1}\label{eq1A}$$

is true for all positive integers $$k$$. Assume there is some such positive integer $$C$$ and set of $$a_i$$. For any prime $$p_j$$ which divides $$C$$ or one of the $$a_i$$, using the $$p$$-adic order function, let

$$b_{j,i} = \nu_{p_{j}}(a_i), \; \; c_j = \nu_{p_{j}}(C) \tag{2}\label{eq2A}$$

We'll prove using induction, with the harmonic numbers (i.e., $$H_1 = 1$$ and $$H_{n+1} = H_n + \frac{1}{n+1} \; \forall \; n \ge 1$$), for all $$k \ge 1$$ that

$$kb_{j,k+1} \le kc_j + \sum_{i=1}^{k}b_{j,i} \le kH_{k}c_j + kb_{j,1} \; \; \to \; \; b_{j,k+1} \le H_{k}c_j + b_{j,1} \tag{3}\label{eq3A}$$

For $$k = 1$$, we get from \eqref{eq1A} and \eqref{eq2A} that

$$b_{j,2} \le c_{j} + b_{j,1} = H_{1}c_{j} + b_{j,1} \tag{4}\label{eq4A}$$

so \eqref{eq3A} works in the base case since $$H_1 = 1$$. Assume \eqref{eq3A} is true for $$k = n$$ for some $$n \ge 1$$. Then, for $$k = n + 1$$, \eqref{eq1A}, \eqref{eq2A} and \eqref{eq3A} give that

\begin{aligned} (n+1)b_{j,n+2} & \le (n + 1)c_j + \sum_{i=1}^{n+1}b_{j,i} \\ & = \left(nc_j + \sum_{i=1}^{n}b_{j,i}\right) + c_j + b_{j,n+1} \\ & \le (nH_{n}c_j + nb_{j,1}) + c_j + (H_{n}c_j + b_{j,1}) \\ & = ((n+1)H_{n} + 1)c_j + (n + 1)b_{j,1} \\ & = (n + 1)\left(H_{n} + \frac{1}{n+1}\right)c_j + (n + 1)b_{j,1} \end{aligned}\tag{5}\label{eq5A}

Dividing both sides of \eqref{eq5A} by $$n+1$$ gives

$$b_{n+2} \le \left(H_{n} + \frac{1}{n+1}\right)c_j + b_{j,1} = H_{n+1}c_j + b_{j,1} \tag{6}\label{eq6A}$$

so \eqref{eq3A} holds for $$k = n + 1$$ as well. Thus, by induction, \eqref{eq3A} holds for all positive integers $$k$$.

Note \eqref{eq3A} shows all prime factors of any $$a_i$$ must divide $$Ca_1$$, so there are only a finite number of them, say some $$r \ge 1$$. Also, the harmonic numbers being a strictly increasing sequence means that $$b_{j,i} \le H_kc_j + b_{j,1}$$ for all $$1 \le i \le k + 1$$. Thus, considering all of the $$r$$ distinct prime factors, we have that all $$a_i$$ for $$1 \le i \le k + 1$$ must be distinct positive factors of

$$M = \prod_{j=1}^{r}p_j^{e_j}, \; e_j = \lfloor H_{k}c_{j} \rfloor + b_{j,1} \tag{7}\label{eq7A}$$

Thus, using the number-of-divisors function $$\sigma_0()$$, the harmonic number upper bound of $$H_k \le \ln(k) + 1$$ and $$c_{m} = \max(c_{1} + 1, \ldots, c_{r} + 1)$$, we get for large enough $$k$$ (i.e., where $$\ln k \gt \max(c_{1} + b_{1,1} + 1, \ldots, c_{r} + b_{r,1} + 1)$$) that

\begin{aligned} k + 1 & \le \sigma_0(M) \\ & = \prod_{j=1}^{r}(\lfloor H_{k}c_{j} \rfloor + b_{j,1} + 1) \\ & \le \prod_{j=1}^{r}(c_{j}\ln{k} + c_{j} + b_{j,1} + 1) \\ & \lt \prod_{j=1}^{r}((c_{j} + 1)\ln{k}) \\ & \le \prod_{j=1}^{r}(c_{m}\ln{k}) \\ & = c_{m}^{r}(\ln{k})^{r} \end{aligned}\tag{8}\label{eq8A}

With $$k = e^x$$, we get $$k \to \infty$$ means $$x \to \infty$$. Also, \eqref{eq8A} becomes

$$e^x + 1 \le c_{m}^{r}x^{r} \tag{9}\label{eq9A}$$

However, exponential functions grow faster than any polynomial (e.g., as can been seen here by using L'Hôpital's rule $$r$$ times to get $$\lim_{x \to \infty}\frac{e^x + 1}{c_{m}^{r}x^{r}} = \lim_{x \to \infty}\frac{e^x}{c_{m}^{r}(r!)} = \infty$$, or by using a post like Proving exponential is growing faster than polynomial), so \eqref{eq9A} can't hold for large enough $$x$$ and, thus, large enough $$k$$.

This contradiction concludes proving that \eqref{eq1A} can't always be true.