Solve $11^x ≡ 21 \pmod {71}$
I am supposed to create a chart with $j$, $11^j$ and $21\cdot 11^{-9j}$ I have filled out the first 2 rows, but can not figure out the 3rd row.
This is what I have so far:
$j = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10$
$11^j \mod {71} = 1, 11, 50, 53, 15, 23, 40, 14, 12, 61, 32$.
I do have that at $j=0$,
$21\cdot 11^{-9j}\mod {71} = 21$,
But I don't have anything further.
Even one example on how to simplify $21\cdot 11^{-9j} \mod {71}$ would be very helpful.
Using your table of the residues of $11^j$ for $0 \le j \le 10$, since $21 \equiv -50 \equiv -(11^2) \pmod{71}$, you just need to determine the value of $y$ where $11^y \equiv -1 \pmod{71}$. You can see that $11^{35} \equiv \left(11^7\right)^5 \equiv 14^5 \equiv -1 \pmod{71}$, giving $y = 35$. Note one way to guess this is also that since by Fermat's little theorem, $11^{70} \equiv 1 \pmod{71}$, then $11^{35} \equiv \pm 1 \pmod{71}$, with it being $-1$ in this case.
Thus, you have $11^{37} \equiv 11^{35} \times 11^{2} \equiv (-1)(-21) \equiv 21 \pmod{71}$, giving that $x = 70n + 37$, for any non-negative integer $n$.
Hint: $71$ is a prime and any natural number of the form $x = 70n + 37$ is a solution of $11^x ≡ 21 \pmod {71}$
Maybe a hint, your start stands, now notice $11^9=61 \pmod{71}$ easily $11^{-9}=7\pmod{71}$. You try $x$ for all values to get the desired $21$ residue. The power order of $11$ in $\mathbb{Z}/71\mathbb{Z} is 70$.
