This is easy if you know LTE

(1): $v_2{(a^n-1)}=v_2(a-1)+v_2(a+1)+v_2(n)-1$ for $2|a-1$ and n even, so all even n work. Clearly, odd n do not work, just from considering $a=3$ (2): $v_2{(a^n-1)}=v_2(a-1)+v_2(a+1)+v_2(n)-1$. We know $v_2(a-1)+v_2(a+1)\ge3$, so $v_2(n)\ge2015$, and $n=k\cdot 2^{2015}$

Part(1) use lifting for exponent lemma and $2|a-1$ Then $\nu_2(a^n-1)=\nu_2(a-1)+\nu_2(a+1)+\nu_2(n)-1$ Since $4|a^n-1$ So $\nu_2(n)\geq1$ Then all even $n$ work Part(2) use lifting for exponent lemma and $2|a-1$ Then $\nu_2(a^n-1)=\nu_2(a-1)+\nu_2(a+1)+\nu_2(n)-1$ Since $2^{2017}|a^n-1$ and $\nu_2(a-1)+\nu_2(a+1)\ge3$ So $\nu_2(n)\geq2015$ Then $n= 2^{2015}k$

lifeisgood03 wrote: (1): $v_2{(a^n-1)}=v_2(a-1)+v_2(a+1)+v_2(n)-1$ for $2|a-1$ and n even, so all even n work. Clearly, odd n do not work, just from considering $a=3$ (2): $v_2{(a^n-1)}=v_2(a-1)+v_2(a+1)+v_2(n)-1$. We know $v_2(a-1)+v_2(a+1)\ge3$, so $v_2(n)\ge2015$, and $n=k\cdot 2^{2015}$ $v_2(a-1)+v_2(a+1)\ge3$ is applied in (2). Why isn't it applicable in (1)? Because if it were, then the answer would have been all positive integers n.

lemon94 wrote: lifeisgood03 wrote: (1): $v_2{(a^n-1)}=v_2(a-1)+v_2(a+1)+v_2(n)-1$ for $2|a-1$ and n even, so all even n work. Clearly, odd n do not work, just from considering $a=3$ (2): $v_2{(a^n-1)}=v_2(a-1)+v_2(a+1)+v_2(n)-1$. We know $v_2(a-1)+v_2(a+1)\ge3$, so $v_2(n)\ge2015$, and $n=k\cdot 2^{2015}$ $v_2(a-1)+v_2(a+1)\ge3$ is applied in (2). Why isn't it applicable in (1)? Because if it were, then the answer would have been all positive integers n. The LTE statement for $v_2$ only applies for even $n$.

Any Proof without LTE?

(1) We need to have that $a^n \equiv 1 \mod{4}$, since $4$ is a number of the form $p^k$, we can pick $a$ to be a primitive root of $4$ such that the order of $a$ mod $4$ is $\varphi(4)=2$. Thus all $2|n$ works. (2) From part 1 we know that $n$ must be even, so we can apply LTE on $a^n-1$ here. We require $$\nu_2(a^n-1)=\nu_2(a-1)+\nu_2(a+1)+\nu_2(n)-1 \geq 2017$$Note that $\nu_2(a-1)+\nu_2(a+1) \geq 3$ as $2k, 2(k+1)$ cannot have both $k, k+1$ be odd numbers. Thus $\nu_2(n) \geq 2015$ so all $n$ such that $2^{2015}|n$ works.

Mathmick51 wrote: Any Proof without LTE? i believe induction works aswell