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I have an exercise to solve an equation like in the title. My goal until now is that both $m$ and $n$ are odd, but then I can not continue. Can you help me? Thanks a lot!

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Consider $m^2+4 (\bmod 11):$

$m=0 (\bmod 11) \Rightarrow m^2+4 = 4 (\bmod 11),$
$m=1,10 (\bmod 11) \Rightarrow m^2+4 = 5 (\bmod 11),$
$m=2, 9 (\bmod 11) \Rightarrow m^2 +4= 8 (\bmod 11),$
$m=3, 8 (\bmod 11) \Rightarrow m^2+4 = 2 (\bmod 11),$
$m=4, 7 (\bmod 11) \Rightarrow m^2+4 = 9 (\bmod 11),$
$m=5, 6 (\bmod 11) \Rightarrow m^2+4 = 7 (\bmod 11).$

But $n^5 (\bmod 11)$:

$n=0 (\bmod 11) \Rightarrow n^5=0 (\bmod 11),$
$n=1,3,4,5,9 (\bmod 11) \Rightarrow n^5=1 (\bmod 11),$
$n=2,6,7,8,10 (\bmod 11) \Rightarrow n^5=10 (\bmod 11).$

So, $n^5\ne m^2+4 (\bmod 11)$, and $n^5\ne m^2+4$, if $n,m\in \mathbb{N}.$

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  • $\begingroup$ @mapping, You are welcome. $\endgroup$ – Oleg567 Sep 7 '15 at 14:02

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