\(2^{6n}=8^{2n}\equiv1\left(mod7\right)\Rightarrow2^{6n}=7k+1\)
\(\Rightarrow2^{6n+2}=4\left(7k+1\right)=28k+4\)
\(\Rightarrow C=2^{28k+4}+13\)
Mặt khác theo định lý Fermat nhỏ:
\(\left(2;29\right)=1\Rightarrow2^{28}-1⋮29\Rightarrow2^{28}\equiv1\left(mod29\right)\)
\(\Rightarrow2^{28k}\equiv1\left(mod29\right)\Rightarrow2^{28k+4}=16.2^{28k}\equiv16\left(mod29\right)\)
\(\Rightarrow2^{28k+4}+13⋮29\)
Hay \(C⋮29\Rightarrow C\) là hợp số