Does
$$a\uparrow \uparrow (n+1) - a\uparrow \uparrow n$$
divide
$$a\uparrow \uparrow (n+2) - a\uparrow \uparrow (n+1)$$
for all $a,n \ge2$ ?
The case n = 0 is easy :
$$a^a-a=a(a^{a-1}-1)$$
and $a-1$ divides $a^{a-1}-1$
The case n = 1
$$a^{a^a}-a^a = a^a (a^{a^a-a}-1)=a^a(a^{a(a^{a-1}-1)}-1)$$
$$a^a-a = a(a^{a-1}-1)$$
and since $a-1$ divides $a^{a-1}-1$, the case is completed.
- Is my proof correct for the cases n=0 and n=1 ?
- Can this process be continued to enable an induction proof ?
Motivation :
A consequence of this divisibility would be :
If
$$a \uparrow \uparrow n \equiv a \uparrow \uparrow(n+1) \ mod\ (\ m\ ) $$
for some $a,n,m \ge2$ , then
$$a \uparrow \uparrow n \equiv a \uparrow \uparrow k \ mod\ (\ m\ ) $$
for all $k \ge n$
Yes, it is correct.
Sure. Let's write $T(a,n) = a \uparrow\uparrow n$. Then
$$\begin{align} T(a,n+2) - T(a,n+1) &= a^{T(a,n+1)} - a^{T(a,n)}\\ &= a^{T(a,n)}\left(a^{T(a,n+1)-T(a,n)}-1\right). \end{align}$$
The induction hypothesis gives $T(a,n+1) - T(a,n) = k\bigl(T(a,n) - T(a,n-1)\bigr)$, and monotonicity gives $T(a,n-1) \leqslant T(a,n)$, whence
$$\begin{align} \frac{T(a,n+2)-T(a,n+1)}{T(a,n+1)-T(a,n)} &= \frac{a^{T(a,n)}\left(a^{T(a,n+1)-T(a,n)}-1\right)}{a^{T(a,n-1)} \bigl(a^{T(a,n)-T(a,n-1)}-1\bigr)}\\ &= a^{T(a,n) - T(a,n-1)}\cdot\frac{a^{k\bigl(T(a,n) - T(a,n-1)\bigr)}-1}{a^{T(a,n)-T(a,n-1)}-1} \end{align}$$
is recognised as the product of two integers, hence an integer.