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Can you find integer values for $p$ and $q$ with $p,q\in\{0,1,2,\ldots,9\}$ so that the following equation holds?

$p^q - q^p = 1844$

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closed as off-topic by Milo Brandt, Deusovi Aug 21 '16 at 4:09

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  • 3
    $\begingroup$ None of the answers give any explanation beyond just the answer, which is trivial to find via computer (or paper and pencil, if you're dedicated). If that is the intended solution (i.e. there is no way to leverage particular properties of $1844$ for a solution), I don't think this is a puzzle. $\endgroup$ – Milo Brandt Aug 20 '16 at 23:42
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I think

p=1845 and q=1 works well EDIT: no longer applicable after new criteria introduced in the question
p=3 and q=7 too

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  • $\begingroup$ 13 seconds faster than me. :) $\endgroup$ – Maria Deleva Aug 20 '16 at 7:27
  • $\begingroup$ Right answer 3 and 7 $\endgroup$ – Smart Aug 20 '16 at 7:31
  • $\begingroup$ But what were the steps you used to get this? $\endgroup$ – manshu Aug 20 '16 at 18:51
  • $\begingroup$ @manshu The case of $p=1845$ and $q=1$ is trivial. After it was stated that we should look for $p$ and $q$ in the range of one-digit numbers, it was just simple brute force for $2\le{}p<q\le9$, where $p$ and $q$ have the same parity. Later by programming I checked for solutions where $\mathrm{max}(p,q)<10000$, but did not find anything else. $\endgroup$ – elias Aug 20 '16 at 18:54
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A non trivial solution

3^7 - 7^3 = 1844

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7
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One possible answer

p = 1845 and q = 1

Because

1845^1 = 1845 1^1845 = 1 1845 - 1 = 1844

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  • $\begingroup$ Sorry I forgot some description. $\endgroup$ – Smart Aug 20 '16 at 7:30
  • $\begingroup$ Well try I upvoted +1 $\endgroup$ – Smart Aug 20 '16 at 7:32

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