The Question is
$$ N=12={ 2 }^{ 2 }+{ 2 }^{ 3 }\\ { M }^{ 2 }\equiv 51(mod\quad 59)\\ What\quad is\quad { M }^{ 12 }(mod\quad 59)? $$
The solution is in the book says its 7. I am not sure what I am supposed to do with N, in relation to the question. Thank you for any insight.
The point of $N$ is to show that if $M^2 \equiv 51 \pmod {59}$, then $$M^{12} = M^{2^2 + 2^3} = M^{2^2} M^{2^3} = (M^2)^2 ((M^2)^2)^2 \equiv (51)^2 ((51)^2)^2 \pmod {59}.$$ Of course, it is easier to write $$M^2 \equiv -8 \pmod {59}$$ so that the squaring operation on $M^2$ is a little less computational.