<s>[INST] 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.
 [/INST] 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.


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*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 $16+16+14+1=47 \equiv 0 \pmod {47}$.
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