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Source. Proposition 3, PDF p. 4 of arXiv:2601.10296v2 (its proof ends on p. 5).

Conventions

For integers $u,v,L$ with $L\geq1$, write

$$S(u,v,L)=\{(m,n)\in\mathbb Z^2:mv\equiv nu\pmod L\}.$$

Two triples are equivalent when their $S$-sets agree. A triple $(u,v,L)$ is semi-reduced when $u$ divides $L$, $\gcd(u,v)=1$, and $1\leq v\leq L$; it is reduced when $v$ is the least positive value among the equivalent semi-reduced triples.

Statement

Let $p$ be a prime not dividing the integers $a$ and $b$. There is a unique reduced triple $(u,v,L)$ such that

$$\{(m,n)\in\mathbb Z_{\geq0}^2:a^m\equiv b^n\pmod p\} =S(u,v,L)\cap\mathbb Z_{\geq0}^2.$$

More precisely, if

$$r=\operatorname{ord}_p(a),\qquad s=\operatorname{ord}_p(b),\qquad L=[r,s],$$

where $[r,s]$ denotes the least common multiple, then

$$u=\frac{L}{s} \qquad\text{and}\qquad \gcd(v,L)=\frac{L}{r}.$$

Proof pointer. The source derives the result from Proposition 2, which chooses a common residue of order $L$. The proof was not reconstructed or independently checked here.

No exact numbered Erdős-problem relationship is assigned here.