parmenides51 wrote: Determine whether exists positive integer $n$ such that the number $A=8^n+47$ is prime. No?? For $n=4k,3|8^n+47$ For $n=4k+2,3|8^n+47$ For $n=4k+3,13|8^n+47$ For $n=4k+1,5|8^n+47$

Claim : No such positive integer $n$ exists. Proof : $A = 8^n + 47$ Take mod 3 $\implies$ $A \equiv (-1)^n + 2 \equiv 0$ (mod $3$) if $n$ is even . Then that means $n$ has to be odd , say for some positive integer $k$ $n = 2k + 1 \implies A \equiv 8^{2k+1} + 47$ , Now seperate $2$ cases of $k$ being odd and $k$ being even . Suppose $k$ is odd $\implies$ $A \equiv 64^{k}\cdot8 + 47 \equiv (-1)^{k}\cdot8 + 8 \equiv 0$ (mod $13$) Now suppose $k$ is even $\implies$ $A \equiv 64^{k}\cdot8 + 47 \equiv (-1)^{k}\cdot3 + 2 \equiv 0$ (mod $5$) So we are done . $\blacksquare$

parmenides51 wrote: Determine whether exists positive integer $n$ such that the number $A=8^n+47$ is prime. https://artofproblemsolving.com/community/c6h2578627

at least 1 problem comes from 2020 JBMO Shortlist (that's why they were not posted online till 2021 JBMO) I do not know which ...

This is NT1 from the JBMO 2020 shortlist (now available online).

Assume that there is such an integer. Taking $\pmod 3$, we get $(-1)^n + 2 \pmod 3$, so $n$ is odd. Let $n=2a+1$. Then we have $8^{2a+1}+47$ is prime. If $a$ is odd, then taking $\pmod {13}$ produces a contradiction. If $a$ is even, then taking $\pmod 5$ produces a contradiction. But this is impossible, hence our original statement was false and there is no such integer.

steppewolf wrote: This is NT1 from the JBMO 2020 shortlist (now available online). what is link?

I have made a video solution on this problem: https://youtu.be/xluIV7cXu6E

Let's suppose that n is an even number, than $8^n + 47 \equiv 1 + 2 $ (mod $3$) This means that n is odd. No we will do some casework based on the remainder of n with 4. n cannot be of form $4k$ or $4k + 2$, as that would be an even number. 1) $n = 4k + 1$ Playing with simple cases has given me the idea that we should use $mod$ $5$. $8 * 8^{4k} + 47 \equiv 3 * 64^{2k} + 2 \equiv 3 * (-1)^{2k} + 2 \equiv 3 * 1 + 2 \equiv 0$ (mod $5$) *Note that $(-1)^{2k}$ is always $1$ 2) $n = 4k + 3$ Again, playing with small cases has given me the idea to use $mod$ $13$ $8^{4k+3} + 47 \equiv 512 * 8^{4k} + 8 \equiv 5 * 64^{2k} + 8 \equiv 5 * 12^{2k} + 8 \equiv 5 * (-1)^{2k} + 8 \equiv 13 \equiv 0$ (mod $13$). Now, usign all these caseworks, we can conclude that $8^n + 47$ is always a composite number.