# 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)