https://artofproblemsolving.com/community/c6h2609453_8n47_is_prime

This problem was proposed by Cyprus

If n=4k,4k+2 then this expression is equal 0 mod 3.If n=4k+1,then expression is 0 mod 5.If n=4k+3 then expression is 0 mod 13.I think this is very easy problem for JBMO and there is nothing difficult to understand

From mod 3 we get that n should be odd (2k+1) and from mod 5 we get that n=4k+1 is not valid (then 8^n +47 will be divisible by 5) so the only solution has to be n=4k+3, but 8^(4k+3) is 5 mod 13 and 47 is -5 mod 13, so then 8^n+47 will be divisible by 13. In conclusion there are no solutions

No. Note that $8^n+47\equiv 2^n + 2\pmod{3}$, so we must have $n$ odd. Also, we have $8^n+47\equiv 3^n+2\pmod{5}$, and hence $n$ cannot be equivalent to $1$ modulo $4$. Let $n = 4n_0+3$. Then, $8^n+47\equiv 8^{4n_0}\cdot8^3 -5 \equiv 512-5\equiv 0 \pmod{13}$. Since $8^n+47>3,5,13$, we are done.