Ta có: \(2005\equiv-1\left(mod2006\right)\)
\(\Rightarrow2005^{2007}\equiv-1\left(mod2006\right)\)
Lại có: \(2007=1\left(mod2006\right)\)
\(\Rightarrow2007^{2005}\equiv1\left(mod2006\right)\)
\(\Rightarrow2005^{2007}+2007^{2005}\equiv0\left(mod2006\right)\)
Vậy \(2005^{2007}+2007^{2005}⋮2006\left(đpcm\right)\)
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