TITLE: How can I solve the functional equation $ f ( x + 1 ) + 1 = f \big( f ( x ) + 1 \big) $?
QUESTION [6 upvotes]: Let $ \mathbb N = \{ 0 , 1 , 2 , \dots \} $. Find all the functions $ f : \mathbb N \to \mathbb N $ such that
$$ f ( x + 1 ) + 1 = f \big( f ( x ) + 1 \big) $$
for all $ x \in \mathbb N $.


I noticed, while looking for injectivity, that
$$ \forall ( x , y , n ) \in \mathbb N ^ 3 : f ( x ) = f ( y ) \implies f ( x + n ) = f ( y + n ) \text . $$
Then I found out that $ f ( 0 ) \ne 0 $, because if it's the case, by letting $ x = 1 $ we'll have
$$ f ( 1 ) + 1 = f \big( f ( 0 ) + 1 \big) \implies 1 = 0 \text . $$

Is there any method to solve this?

REPLY [2 votes]: Theorem 1: The unique function $f:\mathbb{N}\to\mathbb{N}$ satisfying
$$f(x+1)+1=f(f(x)+1)$$
for all $x\in\mathbb{N}$ is given by $f(x)=x+1$ for all $x\in \mathbb{N}$.
Before we can prove this, we need some Lemmas. The proof of Theorem $1$ can be found at the very bottom of this post.

In what follows, let $f:\mathbb{N}\to\mathbb{N}$ be any function satisfying the functional equation.
Lemma 2: Suppose $x,k\in\mathbb{N}$ with $k\ge 1$ and $f(x)=f(x+k)$. Then for all $a,b\in\mathbb{N}$ with $a,b\ge x$ and $a\equiv b\pmod k$, we have $f(a)=f(b)$.
Proof: Note that
$$f(x+1)+1=f(f(x)+1)=f(f(x+k)+1)=f(x+k+1)+1,$$
so $f(x+1)=f(x+k+1)$. With induction on $n$ we conclude that $f(x+n)=f(x+k+n)$ for all $n\in\mathbb{N}$.
Now it suffices to prove that $f(x+mk)=f(x)$ for all $m\in\mathbb{N}$. We proceed by induction on $m$. The case $m=0$ is trivial and the case $m=1$ is given. Let $m\ge 1$ and assume the claim is true for all integers $0,\ldots,m-1$. Then $f(x+(m-1)k)=f(x+mk)$, so putting $n=k$ in the previous identity, we find that
$$f(x+(m+1)k)=f(x+mk)=f(x).$$
This completes the proof by induction.
We now know that $f(x+mk+n)=f(x+n)$ for all $m,n\in\mathbb{N}$, which is the lemma. $\square$
Corollary 3: If $f$ is not injective, then $f$ is bounded.
Proof: If $f(x)=f(x+k)$ for some $x,k\in\mathbb{N}$ and $k\ge 1$, then for all $y\ge x$, the value of $f(y)$ depends only on $y\pmod k$ by Lemma $2$. Therefore, $f$ only takes finitely many values, and must be bounded. $\square$
Lemma 4: $f$ is unbounded and injective.
Proof: By Corollary $3$, it suffices to show that $f$ is unbounded. Assume for the sake of contradiction that $f$ is bounded. Let $M$ be its maximum and let $y\in\mathbb{N}$ with $f(y)=M$.
If $y\ge 1$, then there exists an $x\in\mathbb{N}$ with $y=x+1$, whence $f(f(x)+1) > f(x+1)=f(y)=M$; a contradiction. Therefore, $f(0)=M$ and $f(y)<M$ for all $y\ge 1$.
Let $M_2$ be the maximum of $f$ on $\mathbb{N}_{\ge 1}$, and let $x\in\mathbb{N}$ with $f(x+1)=M_2$, then $f(f(x)+1)>f(x+1)=M_2$, whence $f(f(x)+1)=M$ and $f(x)+1=0$; a contradiction. $\square$
Lemma 5: For all $x,n\in\mathbb{N}$, we have

$f\left(f^{\circ n}(x)+1\right)=f(x+1)+n$.
$f(x+n)+1=f^{\circ n}(f(x)+ 1)$

Proof: Induction on $n$.
Lemma 6 The sequence $0,f(0),f^{\circ2}(0),\ldots$ is a permutation of $\mathbb{N}$, and we have $\operatorname{im}(f)=\mathbb{N}_{\ge 1}$.
Proof: Let $m$ be the minimal value of $f$ on $\mathbb{N}_{\ge 1}$ and let $x_0\in\mathbb{N}$ with $f(x_0+1)=m$. Let $x\in\mathbb{N}$ and write $f(x+1)=f(x_0+1)+n$ for some $n\in\mathbb{N}$, then Lemma $5$ gives
$$f(x+1)=f(x_0+1)+1=f\left(f^{\circ n}(x_0)+1\right)$$
By injectivity, $x=f^{\circ n}(x_0)$. Hence $\mathbb{N}\setminus\{x_0\}\subseteq\operatorname{im}(f)$.
Suppose that $x_0=f(y)$ for some $y\in\mathbb{N}$, then $y=f^{\circ k}(x_0)$ for some $k\ge 0$, so $x_0=f^{\circ(k+1)}(x_0)$. This means that $\#\operatorname{im}(f)=k$, which means $f$ is bounded. This contradicts Lemma $4$, so $x_0\not\in\operatorname{im}(f)$. We conclude that $\operatorname{im}(f)=\mathbb{N}\setminus\{x_0\}$.
However, the first identity in Lemma $5$ implies that $\operatorname{im}(f)=\mathbb{N}_{\ge m}\cup \{f(0)\}$. If $x_0=0$ we are done. Otherwise, $m=2$ and $\{0,1\}=\{f(0),x_0\}$. Because $f(0)\neq 0$ it follows that $x_0=0$. $\square$
Corollary 7 For all $k\in\mathbb{N}_{\ge 1}$, the function $f^{\circ k}$ has no fixed points.

Proof of Theorem 1: Define a new function $g:\mathbb{N}\to\mathbb{N}$ by $g(n):=f(f(n)+2)$. Now, Lemma $5$ gives that for all $n\in\mathbb{N}$,
$$
g(n) = f\left(f(n)+1+1\right) = f\left(f^{\circ n}(f(0)+1)+1\right) = f(f(0)+2)+n=g(0)+n.
$$
Next, with Lemma $6$ and some abuse of notation,
$$
\mathbb{N}_{\ge g(0)} = g(\mathbb{N}) = f\left(f(\mathbb{N})+2\right) = f(\mathbb{N}_{\ge 3}) = \mathbb{N}_{\ge 1}\setminus\{f(0),f(1),f(2)\}.
$$
Hence, $g(0)=4$ and $\{f(0),f(1),f(3)\}=\{1,2,3\}$.
We now prove that $f(0)=1$ and $f(1)=2$. First, if $f(0)=2$, then $4=g(0)=f(f(0)+2)=f(4)$, but $f$ has no fixed points, so $f(0)\neq 2$. Second, if $f(0)=3$, then $f(1)=2$ and $f(2)=1$, because $f$ has no fixed points, but this implies $f^{\circ 2}(1)=1$, even though $f^{\circ 2}$ has no fixed points either. We conclude that $f(0)=1$. Now, if $f(1)\neq 2$, then $f(1)=3$ and $f(2)=2$, which gives a contradiction because $f$ has no fixed points. We conclude that $f(1)=2$.
Finally, assume that the set $S:=\{x\in\mathbb{N}:f(x)\neq x+1\}$ is non-empty. Let $x\in S$ be minimal, then $x\ge 2$. By Lemma $6$, there exists some $y\in \mathbb{N}$ with $f(y)=x+1$. Now, note that
$$
f(y) = f(x-1)+1 = f(f(x-2)+1),
$$
so $y=f(x-2)+1=(x-2)+1+1=x$, by injectivity. This is a contradiction, so we conclude that $S$ is empty. Q.E.D.