For which $n>0$ does $x^{2^n} \equiv 7 (mod \ 9)$ have a solution?
It might be useful to start $x^{2^n} \equiv 16 (mod \ 9)$ but how should one proceed?
Any hints would be appreciated. Thanks!
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this communityFor which $n>0$ does $x^{2^n} \equiv 7 (mod \ 9)$ have a solution?
It might be useful to start $x^{2^n} \equiv 16 (mod \ 9)$ but how should one proceed?
Any hints would be appreciated. Thanks!
In general, if $x^2\equiv a\pmod 9$ has a solution (with $a\not\equiv 0\pmod 3$) then $x^2\equiv -a\pmod 9$ has no solution. Therefore we need only follow one "successful" branch:
Alternatively, if we repeatedly square $7$, we obtain $$ 7, 4, 7, 4, 7, \ldots$$ In other words, $$ 7^{2^n}\equiv 7\pmod 9\qquad\text{if }2\mid n$$ $$ 4^{2^n}\equiv 7\pmod 9\qquad\text{if }2\not\mid n$$