Problem 7 *

Screenshot 2020-07-07 at 19.53.28 Screenshot 2020-07-07 at 19.53.55

\therefore 17|p and p is prime so we can conclude that p=17
17^b=(19-2)^b=19^b+\binom{b}{1}19^{b-1}\cdot (-2)+...+\binom{b}{b-1}19\cdot (-2)^{b-1}+(-2)^b\equiv (-2)^b(\mod19)

Rewriting our equation (\mod19):


This implies that 2^a+(-2)^b\equiv0(\mod19) which is clearly only true if 2^a+(-2)^b=0\Rightarrow a=b and a,b are odd.

\hspace{0.6cm}=2\underbrace{(19^{a-1}+19^{a-2}\cdot 17+...+17^{a-1})}_{sum\;of\;a\;odd\; terms\;\Rightarrow \;odd\; sum\; unless\; a=1}

For all a>1,2^a has no odd factors (other than 1) hence it must be the case that 19^{a-1}+...+17^{a-1} is even \therefore a=1. This gives us only one triple (a,b,p)=(1,1,17).