Prove there is only one solution to the Diophantine equation $p^n - p = q^m - q$ where $p$ and $q$ are odd primes $p\gt q$

Question

Consider numbers of the form $p^n - p$ where $p>2$ is a prime and $n>1 \in \mathbb{Z}$. How many of these have a unique representation? $2184$ can be written in this form $2$ ways, $3^7-3, 13^3-13$, can any other numbers?

The OEIS sequence entry A057896 consists of numbers which can be expressed $m^k-m$ more than one way. The number $2184$ is the only one listed which has $m$ prime for both solutions and a computational bound of $10^{24}$ is given under which no other solutions have been found. The related sequence A308324 is a subsequence of A057896 which allows only prime $m$.

The question is closely related to the the Pillai Conjecture since uniqueness would imply there are no other solutions (except $3,13$) to the Diophantine Equation $p^n - q^m = p - q$. In 2001 Bennett showed that for fixed p and q this equation has at most two solutions (no triple representations) and further conjectured that the general equation (allowing $p$ and $q$ to be composite) has only the 8 solutions which he presents. For only one of those solutions $(3,13)$ is $p$ and $q$ both prime.

If the conjecture made by Bennett is correct this is the only solution to this equation. See his paper On Some Exponential Equations of S. S. Pillai for more details.

Answer 1

More Progress

This idea is an attempt to generalize the proof of a special case by @starfall. This approach does not imply there exists a solution if the following two cases don't hold. It only provides some cases where solutions cannot exist.

Case 1

Let $k=ord_p q$ and $c=(q^{k}-1)/p$.

If $c\equiv 0 \pmod p$ then a solution cannot exist since $p(p^{n-1}-1)=q(q^{m-1}-1)$ implies p divides $(q^{m-1}-1)$ only once.

An example for this case is $(p,q,k,c)=(11,3,5,22)$.

If we apply the same logic symmetrically, swapping p for q we have the examples $$\{(17,3,2,96),(7,5,4,480),(19,3,1,6),(19,7,6,6720840),(19,13,12,170254993774320)\}.$$

Case 2

Further, if we let $j=ord_c p$ there can also be no solution if $p^j \equiv 1 \pmod {q^2}$ since q can divide $p^{n-1}-1$ only once. Some examples of this case are:

$$(p,q,k,c,j)= \{(7,3,6,104,12) ,(11,5,5,284,70) ,(13,7,12,1064714400,2520) ,(17,3,16,2532160,960) ,(19,3,18,20390552,756) ,(19,5,9,102796,4140) ,(19,13,18,5918705629050389112,104628420) ,(19,17,9,6241467184,21601152)\}$$

Applying the same logic symmetrically we have the example $(17,13,6,1856736,1224).$

Instead of c we can also use the squarefree part of c and arrive at the same result.

Considering these two cases for $3\le q<p\le 19$ only leaves the possible solutions: $$(p,q)=\{(5,3),(11,7),(13,3),(13,5),(13,11),(17,5),(17,7),(17,11),(19,11)\}.$$

Regardless, there do not seem to be any solutions for $3\le q<p\le 19$ for $n \lt 10^9$ using a search based on the fact that $m=floor(n \ln(p)/\ln(q))$ and naive modular exponentiation. This is consistent with Theorem 1.5 from Bennett which states there can be no additional solutions for $p-q \le 100$.

Answer 2

This is a for solution n = 3 and m = (6k+1) for a positive integer k.

(1) ... $p^3-p=q^{6k+1}-q$ can be factored as

(2) ... $p(p-1)(p+1) = q(q^k-1)(q^k+1)(q^{2k}-q^k+q)(q^{2k}+q^k+1)$

From (1) we can get (3) ... $p^3 > q^{6k+1}$ which would also imply $p > q^{2k}$

From (2) we can see that p must divide either $(q^{2k}-q^k+q)$ or $(q^{2k}+q^k+1)$

That would mean that (4) ... $p \le (q^{2k}+q^k+1)$.

Combining (3) and (4) we get: $q^{6k+1} < p^3 \le (q^{2k}+q^k+1)^3$

Which would imply: $q^{6k+1} < (q^{2k}+q^k+1)^3$

The only solution to this inequality is $q = 3, k=1$. We could do induction on both $q$ and $k$ to show this but it gets unwieldy

This method would work for other Superior highly composite numbers for m. ex. $m = 12k+1$.

Unfortunately I couldn't find a way to generalize it.