TITLE: Give an inductive proof for the following divisibility statement.
QUESTION [7 upvotes]: I wish to prove that 
$$19 \mid 2^{2^{6n+2}}+3 \ \ \text{for all} \ \  n\in\mathbb{N}\cup\{0\}$$
This is problem 5 in the book "250 Problems in Elementary Number Theory" (1970) by Waclaw Sierpinski.
Thank you in advance.

REPLY [2 votes]: We have that $$2^6\equiv 1\pmod{9}\implies 2^{6k}\equiv 1\pmod{9}\implies 2^{6k+2}\equiv 4\pmod{9}$$ from where (since LHS and RHS are even) $$2^{6k+2}\equiv 4\pmod{18}.$$ That is $$2^{6k+2}=18n+4$$ for some nonnegative integer $n.$ Thus
$$2^{2^{6k+2}}=2^{18n+4}\equiv 2^4\pmod{19}$$ where we have used Fermat's theorem ($2^{18}\equiv 1\pmod{19}\implies 2^{18n}\equiv 1\pmod{19}$). Finally, note that
$$2^4+3=19$$ and we are done.