Talk:Power of 2 is Difference between Two Powers

Unsolved, 2nd ed (on D9 Difference of two powers, p. 155-157) writes:

Hugh Edgar asks how many solutions $(m, n)$ does $p^m - q^n = 2^h$ have, for primes $p$ and $q$ and $h$ an integer?

He gives these examples and asked are there others. In the book he cited partial results

A.O. Gelfond, Sur la divisibilite de la difference des puissances de deux nombres entiers par une puissance d'un ideal premier, Mat. Sbornik, 7(1940) 724

Howard Rumset & Edward C. Posner, On a class of exponential equations, Proc. Amer. Math. Soc., 15(1964) 974-978

implying there are finitely many, although it probably refers to the number of $(m, n)$, not $h$.

On a separate note, it is not sufficient to only require the bases $p, q$ to be distinct to eliminate trivial solutions, as:

$2^{4 + 2 n} = \paren {5 \times 2^n}^2 - \paren {3 \times 2^n}^2$

$2^{5 + 2 n} = \paren {9 \times 2^n}^2 - \paren {7 \times 2^n}^2$

I am not sure if $p, q$ being coprime can resolve the issue (of being nontrivial and true), but for $64$ there is no solution for primes $p, q$ up to $p^m, q^n < 10^{18}$, while:

$2^6 = 17^2 - 15^2$

Reference: http://www.sspectra.com/Pillai.txt --RandomUndergrad (talk) 06:59, 28 July 2020 (UTC)

Thank you for clarifying this, I'll give it some proper thought as to how to present this page and redraft it. --prime mover (talk) 08:18, 28 July 2020 (UTC)

Update: on the issue of coprimality, as the identity $2^n = \paren {2^{n - 2} + 1}^2 - \paren {2^{n - 2} - 1}^2$ suggests, this problem is meant for primes only. --RandomUndergrad (talk) 11:24, 3 August 2020 (UTC)