Integer to an integer in number theory ?

Let's prove the existence of a particular solution. Let $n = 2017^{a_1}$ . Then we would need (remember that $\phi(2^k) = 2^{k-1}$ ) $2017^{a_1 2017^{a_1}} - 2017 \equiv 0 \mod 2^{2017} \implies 2017^{a_1 2017^{a_1} - 1} \equiv 1 \mod 2^{2017} \implies a_1 2017^{a_1} - 1 \equiv 0 \mod 2^{2016}$ Now choose $a_1 = 2017^{a_2}$ . Then we would need $2017^{a_2 + 2017^{a_2}} - 1 \equiv 0 \mod 2^{2016} \implies a_2 + 2017^{a_2} \equiv 0 \mod 2^{2015}$ . For this to be possible, $a_2$ would need to be odd so for convenience, remember that $2017 \equiv -k \mod 2^{2015}$ for some $k$ . Thus, we can write the previous congruence as $k^{a_2} - a_2 \equiv 0 \mod 2^{2015}$ . Now, choose $a_2 = k^{a_3}$ . Then $k^{k^{a_3}} - k^{a_3} \equiv 0 \mod 2^{2015} \implies k^{a_3} \left( k^{k^{a_3} - a_3} - 1 \right) \equiv 0 \mod 2^{2015} \implies k^{a_3} - a_3 \equiv 0 \mod 2^{2014}$ Now we have ended in a congruence with the same form as the one we started with. So we can now use the substitution $a_3 = k^{a_4}$ and repeat the entire procedure to get $k^{a_4} - a_4 \equiv 0 \mod 2^{2013}$ . If we do this enough times we will end up with $k^{a_{2016}} - a_{2016} \equiv 0 \mod 2$ and then we can just choose $a_{2016} = 1$ . We can now just substitute back until we find $a_1$ .

So a solution is $n = 2017^{2017^{(2^{2015} - 2017)^{(2^{2015} - 2017})^{...^{2^{2015} - 2017}}}}$

Note: The procedure outlined here and easily be generalized to prove that given $a,b \in \mathbb{N}$ , $a$ odd, there always exist solutions to the congruence $x^x - a \equiv 0 \mod 2^b$