sandu2508 wrote: Let $ p$ be a positive integer. Define the function $ f: \mathbb{N}\to\mathbb{N}$ by $ f(n) = a_1^p + a_2^p + \cdots + a_m^p$, where $ a_1, a_2,\ldots, a_m$ are the decimal digits of $ n$ ($ n = \overline{a_1a_2\ldots a_m}$). Prove that every sequence $ (b_k)^\infty_{k = 0}$ of positive integer that satisfy $ b_{k + 1} = f(b_k)$ for all $ k\in\mathbb{N}$, has a finite number of distinct terms. $ \mathbb{N} = \{1,2,3\ldots\}$ It's easy to see that the equation $ 9^p(1 + \log_{10}(x)) = x$ has two positive roots $ x_1 < x_2$ and that $ 9^p(1 + \log_{10}(x))\le x$ $ \forall x\ge x_2$ Let $ M = \max(b_0,x_2)$ It's easy to show with induction that $ b_n\le M$ $ \forall n\in\mathbb N_0$ : It's obviously true for $ n = 0$ : $ b_0\le M$ (definition of $ M$). If $ b_k\le M$, then : if $ x_2\le b_k\le b_0$, then $ b_{k + 1} = f(b_k)\le 9^p(1 + [\log_{10}(b_k)])\le b_k$ (definition of $ x_2$) and so $ b_{k + 1}\le M$ if $ b_k < x_2$ then $ b_{k + 1} = f(b_k)\le 9^p(1 + [\log_{10}(b_k)])\le 9^p(1 + \log_{10}(x_2)) = x_2 \le M$ Then, the sequence of positive integers has an upper bound and so has a finite number of distinct terms. Q.E.D.

Beautiful solution. Thanks you.