Problem 1111

1.22. Let an odd prime number $p$ be chosen. We will prove that any two numbers from the increasing sequence of even numbers $a_{m}=p 2^{m}+1\left(m \in \mathbf{Z}^{+}\right)$ do not have common divisors greater than two. Indeed, if $m>l \geqslant 0$, then the number

$$\begin{aligned} p^{p 2^{m}}-1=\left(p 2^{m-1}+1\right) & \left(p^{m-1}-1\right)=\ldots \\ & \cdots=\left(p^{2 m-1}+1\right)\left(p 2^{m-2}+1\right) \ldots\left(p^{l}+1\right)\left(p 2^{l}-1\right) \end{aligned}$$

is divisible by $p^{2}+1$, therefore

$$\left(a_{m}, a_{l}\right)=\left(2+\left(p 2^{m}-1\right), p^{2}+1\right)=\left(2, p^{2}+1\right)=2 .$$

Consider the set of those members of the sequence that are divisible by at least one prime number greater than $p$. This set is not empty, because, as a result of the property of the sequence $\left\{a_{m}\right\}$ proven above, only a finite number of its members can lack divisors greater than $p$. Let $a_{m}$ be the smallest number in this set. Then we have

$$h\left(a_{m}\right)>p, h\left(a_{m}-1\right)=h\left(p^{2 m}\right)=p$$

$$h\left(a_{m}-2\right)=h\left(p^{2^{m}}-1\right)=\max \left\{h\left(p^{m-1}+1\right), h\left(p^{m-1}-1\right)\right\}=\ldots$$

$$\ldots=\max \left\{h\left(p^{m-1}+1\right), h\left(p 2^{m-z}+1\right), \ldots, h(p+1), h(p-1)\right\}=$$

$$=\max \left\{h\left(a_{m-1}\right), h\left(a_{m-2}\right), \ldots, h\left(a_{0}\right), h(p-1)\right\}<p,$$

since $h(p-1)<p$ and for each value of $t=0,1, \ldots, m-2$, $m-1$, according to the choice of the number $m$, we have $h\left(a_{l}\right) \leqslant p$, and moreover, $a_{l} \equiv 1(\bmod p)$, from which $h\left(a_{l}\right)<p$. Thus, for a given odd prime number $p$, a number of the form $n=p^{2^{m}}-1$ is specified, satisfying the inequalities

$$h(n)<h(n+1)<h(n+2) .$$

Since different numbers $p$ will yield different values of $n_{\text {; }}$, the set of such values is infinite.