Is $7^{8}+8^{9}+9^{7}+1$ a prime? (no computer usage allowed) 
Prove or disprove that $$7^{8}+8^{9}+9^{7}+1$$ is a prime number, without using a computer.

I tried to transform $n^{n+1}+(n+1)^{n+2}+(n+2)^{n}+1$, unsuccessfully, no useful conclusion.
 A: Trial division mod $47$, as per the hint.
Firstly, for positive integer $a$, observe that $$a \equiv \overbrace{a \mod 50}^{\text{reduced residue}}+3\lfloor a/50\rfloor \pmod {47}.$$  This makes taking mod $47$s much easier.


*

*We compute $7^2=49 \equiv 2 \pmod {47}$.  So $7^8 \equiv 2^4 = 16 \pmod {47}$.

*We compute $8^2=64 \equiv 14+3 = 17 \pmod {47}$.  So $8^4 \equiv 17^2 = 289 \equiv 39+3 \times 5=54 \equiv 4+3=7 \pmod {47}.$  So $8^8 \equiv 7^2 = 49 \equiv 2 \pmod 7 \pmod {47}.$  So $8^9 \equiv 2 \times 8 = 16 \pmod {47}$.

*I happen to have memorized that $9^3=729$ (I used to set my alarm to 7:29am because it is equal to $3^6$).  So $9^3 \equiv 729 = 29+3 \times 14=71 \equiv 21+3=24 \pmod {47}$.  So $9^6 \equiv 24^2=576 \equiv 26+3 \times 11=59 \equiv 9+3=12 \pmod {47}$.  So $9^7=12 \times 9=108=8+3 \times 2=14 \pmod {47}$.
Finally $16+16+14+1=47 \equiv 0 \pmod {47}$.
A: That $\,47\,$ is a factor can be verified by very easy mental arithmetic:
$\begin{eqnarray} 
{\rm mod}\ 47\!:\quad && 1+ \color{#c00}{7^8} +\, 8^9 +\ \color{#0a0}{9^7}\\
\equiv && 1+  \color{#c00}{2^4} + 2^{27} + \color{#0a0}{7\cdot 7^6\cdot 8^7}\quad {\rm by}\ \ \color{#0a0}{9\equiv 7\cdot 8},\ \ \color{#c00}{7^2\equiv 2}\\
\equiv&& 1+ 2^4 + 2^{27} + 7\cdot \color{#c00}{2^3}\cdot 2^{21}\\
\equiv&& 1+ 2^4 + 2^{4} +\ 7\cdot 2\,\equiv\, 0\quad {\rm by}\ \ \color{#c00}2^{23}\equiv (\color{#c00}{7^2})^{23}\equiv 7^{46}\equiv 1,\ \ \rm by\ little \ Fermat\\
\end{eqnarray}$
