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I am interested in finding solutions to the following equation $$x! + !x = a^3$$ where $x$ and $a$ are natural numbers and $!x$ is the subfactorial of $x$. I've found the solutions $x=1$ and $x=3$.

Are there any other solutions? What methods can be used to find out solutions to problems like these without utilizing the hit and trial method?

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  • $\begingroup$ The term natural number is vague. Some people use it to include zero, while others don't. Do you mean "positive integer" or do you mean "non-negative integer"? $\endgroup$ – Joel Reyes Noche Feb 23 '15 at 4:15
  • $\begingroup$ Oh lord; I had initially written whole numbers to specify All natural numbers as well as zero. People suggested I edit to Natural numbers; now you say Natural numbers is vague.(Not being rude; just exasperating.) Yes, positive integer is what I mean (Besides, since zero isn't a solution it doesn't matter anyway whether Natural Numbers here includes zero or not.) $\endgroup$ – Kugelblitz Feb 23 '15 at 4:21
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    $\begingroup$ For what it's worth, the numbers $n!+!n$ are tabulated at oeis.org/A236438 where it is seen that they satisfy the recurrence, $a_n=na_{n-1}+(-1)^n$, $a_0=2$. (ed ajf) $\endgroup$ – Gerry Myerson Feb 24 '15 at 5:12
  • $\begingroup$ @GerryMyerson Thank you for the site; I'll see if I can get more productive/inspired from having a closer look at the recurrence. Probably because I randomly came up with it usukidoll... $\endgroup$ – Kugelblitz Feb 24 '15 at 5:18

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