Ackermann function $A(m, n)$, all nonnegative integer solutions to $A(m, n) = m + n$?

Question

The Ackermann function $A(m, n)$ is given by the recursion$$\begin{cases} A(0, n) \overset{\text{def}}{=} n + 1 \\ A(m + 1, 0) \overset{\text{def}}{=} A(m, 1) \\ A(m + 1, n + 1) \overset{\text{def}}{=} A(m, A(m + 1, n)).\end{cases}$$What are all non-negative integer solutions of the equation $A(m, n) = m + n$?

Answer 1

$A(m,n) \geq n+m+1$ for all $m,n \geq 0$. Proof: by induction to $m$.

Base case: $A(0,n) =n+1 \geq n+1$.

Induction hypothesis (IH1): $A(m,n) \geq m+n+1$ for all $n \geq 0$ and some $m$.

Inductive step: Within the inductive step, we need to do induction to $n$.

Base case: We have that $$A(m+1,0)=A(m,1)\geq m+2 = (m+1)+0+1$$ where the inequality holds by IH1.

Induction hypothesis (IH2): $A(m+1,n)>(m+1)+n+1$ for some $n\geq0$.

Inductive step: We have that \begin{align*} A(m+1,n+1) &=A(m,A(m+1,n)) \\&\geq m+A(m+1,n)+1 \\&\geq m+m+1+n+1+1 \\ &= 2m+n+3 \\&\geq m+n+3 \\&= (m+1)+(n+1)+1 \end{align*}

The first inequality holds because IH1, and the second holds because of IH2. This completes this inductive step, hence $A(m+1,n) \geq (m+1)+n+1$ for all $n \geq 0$.

This also completes the other inductive step.

Hence $A(m,n) \geq n+m+1>m+n$, so there are no solutions.

Answer 2

A simpler proof of no solutions using the fact that the Ackermann function is strictly increasing in both arguments:

$$A(m,n)\ge m+A(0,n)=m+n+1$$

Of course, to prove it is strictly increasing in both arguments requires induction but I think it is a pretty obvious and simple property to prove.