Solving a congruence without Fermat's little theorem

Question

Given $n\in\Bbb N$, what is the least $a>1$ with $a^{2^n}\equiv1\bmod2015$?

Is there a solution not using Fermat's little theorem or the Chinese remainder theorem, any ideia?

Answer 1

The given problem statement can be written like this:- $$a^{2^n}-1=2015x \tag{1.}$$ for some positive integer $x$. If we divide eqation 1 by $a-1$:-
i.e. $$\frac{a^{2^n}-1}{a-1}=\frac{2015x}{a-1} \tag{2.}$$ i.e. $$1+a+a^2+...........+a^{2^n-1}=\frac{5\times13\times 31\times x}{a-1} \tag{3.}$$ - (Geometric progression)
Since L.H.S. of eq. (3) is an integer (because a is an integer),so R.H.S. would also be an integer.According to equation (2.) lower the $x$, lower will be $a$, so we can take $x=1$. Now we can easily see the smallest $(a-1)=5\implies a=6$ which will divide R.H.S. integrally.