$f(2) = f(f(1))+1 $ $f(3)= f(f(f(1))+1 )+1$ $f(4) = f(f(f(f(1))+1 )+1)+1$ $f(5) = f( f(f(f(f(1))+1 )+1)+1)+1$ . . . . . . . $f(10^{10}) = f(10^{10}-1)+1 $ $f(10^{10}) = f(............f(f(1))+1)+1$ $f(10^{10}) = f(...........f(f(1))+1)+1+1+.......)$ $10^{10}-2$ times $f(f(f(..........f(f(1))+1.....+1+1+1)$ $f(x) = (10^{10})k+x$ for (k=integer)

VicKmath7 wrote: Let $S=${$1,2,...,10^{10}$}. Find all functions from $S$ to itself, such that $f(x+1)=f(f(x))+1(mod 10^{10})$ for each $x$ from $S$. $x = 10^{10}$, you get $f(10^{10} + 1)$ in the LHS, which is not defined?

Now it is, check again.