Let $p(n)$ is the biggest prime divisor of $n$. Prove that, exist infinite $n \in N$ satify : $p(n) <p(n+1)<p(n+2)$ Let $p(n)$ denote the biggest prime factor of $n$. Prove that, there are infinite $n \in N$ satisfy : $p(n) <p(n+1)<p(n+2)$ My idea is finding some special consequences like $a_n=2^{2^m+1}$ and $n=a_n$ $p(n+1) \geq 3>2=p(n)$ and $p(n+2)=2(2^{2^m}+1)$ where $2^{2^m}+1$ is Fermat number If there are infinite Fermat number the problem is solved. (but this statement is more difficult than the problem), Thank you for helping, Finally I solved it With same previous idea . Let p are odd prime Exist $k_0 = \inf \\{k \in N| P(p^{2^k}+1) > p\\} ; k_0 < \infty .$ ($P(n)$ same meaning with $p(n)$) By using Lemme : $P(p^{2^{k_0}}+1) \equiv 1 (\mod 2^{k_0+1})$ And then , we prove : $P(p^{2^{k_0}}-1)<P(p^{2^{k_0}})< P(p^{2^{k_0}}+1).$

From page 320 of "On the largest prime factors of $n$ and $n+1$" by Paul Erdos and Carl Pomerance ( _Aequationes Mathematicae_ 17, 1978, pp. 311-321):

> Suppose now $p$ is an odd prime and
>
> $$k_0=\inf\\{k:P(p^{2^k}+1)\gt p\\}$$
>
> (note that $P(p^{2^{k_0}}+1)\equiv 1$ mod$(2^{k_0+1})$, so $k_0\lt\infty$). Then
>
> $$P(p^{2^{k_0}}-1)\lt P(p^{2^{k_0}})\lt P(p^{2^{k_0}}+1)$$

Remark: They go on to say:

> On the other hand, we cannot find infinitely many $n$ for which
>
> $$P(n)\gt P(n+1)\gt P(n+2)$$
>
> but perhaps we overlook a simple proof.