Let us first write an arbitrary natural number. For example, $$2538$$ Add the squares of its digits $$2^2 + 5^2 + 8^2 +3^2 = 102$$ Next, we do the same with the number obtained $$1^2 + 0^2 + 2^2 = 5$$ Proceed in the same way
\begin{align} 5^2&=25\\ 2^2+5^2&=29\\ 2^2+9^2&=85\\ 8^2+5^2&=89\\ \vdots \end{align}
Prove that unless this procedure leads to the number $1$ (in which case the number 1 will of course recur indefinitely, it must lead to the number $145$, and the following cycle will occur again and again: $$145, 42, 20, 4, 16 , 37, 58, 89$$
I asked myself the following questions: ''How big can the sum of squares of digits of a three-digit number be''? ''A four-digit number''? ''Can you show that for all sufficiently large numbers, the sum will always be smaller than the original number''? How to prove it? I have no ideas