True or false? (I don't know.) Every positive integer is the difference of two powers. Examples:
$ 1 = 3^2  2^3 $
$ 2 = 3^3  5^2 $
$ 3 = 2^75^3 $
$ 4 = 2^32^2 = 5^311^2 $
$ 5 = 2^5  3^3 $
$ 6 = ? $
$ 7 = 2^55^2 $
$ 8 =? $
$ 9 = 5^22^4 $
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5$\begingroup$ I don't think this question is as bad as it seems. It is clear that every integer not congruent to $2$ mod $4$ is a difference of two powers (namely two squares). However, for integers congruent to $2$ mod $4$ the answer is not clear to me. $\endgroup$ – GH from MO Apr 11 '18 at 19:11

5$\begingroup$ According to the Wikipedia page on Catalan's conjecture (en.wikipedia.org/wiki/Catalan%27s_conjecture), $6$ is not the difference of two powers. I would be delighted to see a proof of this result. $\endgroup$ – GH from MO Apr 11 '18 at 19:18

4$\begingroup$ @abx: The equation is not $n=a^2b^2$ but $n=a^pb^q$. It is definitely of research level, see the linked Wikipedia page. $\endgroup$ – GH from MO Apr 11 '18 at 19:43

6$\begingroup$ @GerryMyerson: It is a notorious open problem to say anything about the solutions of a single equation $a^pb^q=n$ (where $a,b,p,q$ are all variables). On the other hand, it might be more accessible to show that the equation has no solution for some $n$ (without specifying this $n$), and that would answer the OP's question. $\endgroup$ – GH from MO Apr 11 '18 at 21:36

6$\begingroup$ Well, anyway, I can fill in one of the blanks: $8=2^30^3$. If use of zero is objectionable, then $8=2^42^3$. oeis.org/A074981 tabulates a conjectured list of positive integers not a difference of two powers; it begins 6, 14, 34, 42, 50, 58, 62, 66, 70, 78, 82, 86, 90. $\endgroup$ – Gerry Myerson Apr 11 '18 at 22:49