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Given a prime p and a prime q, is there some $\alpha$ and $\beta$ such that $p^\alpha = q^\beta + 1$ or $q^\beta = p^\alpha + 1$?

I am unable to formalize a proof for this, but it seems to me to be true. My reasoning is that, WLOG say $p > q$ ,take any $q^\gamma$ then $\exists\alpha$ s.t. $p^\alpha < q^\gamma < p^{\alpha+1}$. Thus, the distance from $q^\gamma$ to one of $p^\alpha$ or $p^{\alpha+1}$ is less than $p^\alpha/2$. Now take $q^{\gamma+1}$, the maximum distance this power of $q$ can be from a power of $p$ is no more than $p^{\alpha+1}/2$. And since, $\Pi_{i=0}^\inf (1-1/a^i) = 0$ (since $a = p > 0$), it seems unlikely to miss infinitely

I know this isn't the way I should be thinking about it, because it is a purely probabilistic approach. Is this property guaranteed to hold like in $p = 2, q=5$ or does it only hold in special cases?

Andrew Butler
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Above and beyond what's been mentioned in the comments, you're also overlooking the trivial here: unless one of $p$ or $q$ is $2$, then their powers can never differ by only one — the powers (like the primes themselves) will be odd, and the difference between odd numbers is always an even number.

Steven Stadnicki
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  • Is there any limit as to the smallest $y$ that satisfies $p^\alpha = q^\beta + y$ given any $p$ and $q$ – Andrew Butler Mar 05 '18 at 06:54
  • @AndrewButler The ABC conjecture essentially implies that $p^\alpha \gt (pqy)^{1+\epsilon}$ only finitely often; this means that $y$ essentially always has to be roughly the size of $p^{\alpha-1}q^{-1}$ or so. It's conjectured that for any specific value $c$ there are only finitely many differences of (relatively) prime powers less than $c$. – Steven Stadnicki Mar 05 '18 at 16:58