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Can all positive integers $k$, be written as a difference of two perfect powers $k=a^m-b^n$, with $m,n>1$ and $a,b$ positive integers?

A number is imperfect if it can not, which numbers are imperfect?

What is the asymptotics of the number of imperfect numbers less then $x$, as $x\rightarrow\infty$?

I have proved that all odd numbers are the difference of two squares. How to solve the other cases?

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    A number is the difference of two squares if and only if it is odd or divisible by 4.2012-06-03
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    So we are left with k=2*odd, and m or n above 22012-06-03
  • 6
    The terminology "[perfect number](http://en.wikipedia.org/wiki/Perfect_number)" already has a standard meaning in number theory - you should probably choose a different word.2012-06-03
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    One exception: $4$ is not the difference of two positive squares. It is, however, $2^3-2^2$. To supplement NS's comment: $$\begin{array}{c} (k+1)^2-(k-1)^2=4k; \\ (k+1)^2-k^2=2k+1. \end{array}$$ This covers integers congruent to $0$ or $\pm1$ modulo $4$.2012-06-03

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See http://oeis.org/A074981 and references there.