a) clearly $5|2^{4n+2}+1$ for all $n\in\mathbb{N}$, so we need to find those $n$ for which $13|2^{4n+2}+1$, it is very easy to show, $13|2^a+1\implies a\equiv 0\pmod 6\implies (4n+2)\equiv 0\pmod 6\implies n\equiv 1\pmod 3$. b). clearly, $5| 2^{4n+2}+1$ for all $n\in \mathbb{N}$, so the only possibility is if $2^{4n+2}+1=5$ which does not have a solution in $\mathbb{N}$, hence no solution can be found.

uh sorry to revive this thread but how does the part b solution work since its already divided by 65

legend267 wrote: uh sorry to revive this thread but how does the part b solution work since its already divided by 65 I think that is a mistake. but second part seems a bit hard. the equation becomes: $$A_n = \frac{2^{6(2k+1)}+1}{65}$$which has sometihng to do with binomial theorem? Also, can we factorise by Sophie Germain Identity?

a)$2^{4n+2}+1=16^n*4+1 \equiv 0 \pmod{5}$ $2^{4n+2}+1=16^n*4+1 \equiv 3^n*4+1$ $4*3^n+1 \equiv 0 \pmod{13} \to 3^n \equiv 3 \pmod{ 13} \to n \equiv 1 \pmod {3}$ So $n=3k+1$ b) $A_n=\frac{ 2^{4n+2}+1}{65}=\frac{ 4*(2^n)^4+1}{65}=\frac{(2*2^{2n}-2*2^n+1)(2*2^{2n}+2*2^n+1)}{65}=\frac{(2^{6k+3}-2^{3k+1}+1)(2^{6k+3}+2^{3k+1}+1)}{65}$ So $(2^{6k+3}-2^{3k+1}+1)(2^{6k+3}+2^{3k+1}+1)=65p$, where $p$ is prime. If $k \geq 1$ then $2^{6k+3}-2^{3k+1}+1 \geq 497$, but it means that $p|2^{6k+3}-2^{3k+1}+1 \to 2^{6k+3}+2^{3k+1}+1 \leq 65 \to $ contradiction. For $k=0$ we get $n=1,A_1=1$ is not prime So $A_n$ is not prime for every $n$.