Hello, flash101!

I'm not familar with "using indices", but here's my version of #1 . . .

1.) Find all the solutions of 4^x .= .13 .(mod 17)

Examine consecutive powers of 4:

. . 4^0 .= .1 .(mod 17)

. . 4^1 .= .4

. . 4^2 .= -1

. . 4^3 .= 13

. . 4^4 .= .1

. . 4^5 .= .4

. . 4^6 .= -1

. . 4^7 .= 13

We see that the mod-17 values are cyclic: 1, 4, -1, 13, . . .

. . and the solutions are: .x = 3, 7, 11, 15, . . .

That is: .x .= .4k - 1, for any integerk.