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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 = 2^7-5^3 $
$ 4 = 2^3-2^2 = 5^3-11^2 $
$ 5 = 2^5 - 3^3 $
$ 6 = ? $
$ 7 = 2^5-5^2 $
$ 8 =? $
$ 9 = 5^2-2^4 $

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    $\begingroup$ 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
    Commented Apr 11, 2018 at 19:11
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    $\begingroup$ 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
    Commented Apr 11, 2018 at 19:18
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    $\begingroup$ @abx: The equation is not $n=a^2-b^2$ but $n=a^p-b^q$. It is definitely of research level, see the linked Wikipedia page. $\endgroup$
    – GH from MO
    Commented Apr 11, 2018 at 19:43
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    $\begingroup$ @GerryMyerson: It is a notorious open problem to say anything about the solutions of a single equation $a^p-b^q=n$ (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
    Commented Apr 11, 2018 at 21:36
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    $\begingroup$ Well, anyway, I can fill in one of the blanks: $8=2^3-0^3$. If use of zero is objectionable, then $8=2^4-2^3$. 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
    Commented Apr 11, 2018 at 22:49

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