Need help of finding residue

Question

I need to find residues of $$7^n+11^n\pmod{19}$$ I have no idea where to start... n is a natural number.

Answer 1

Hint: For $n=1,\ldots,6$ the remainder is $-1,-1,2,-1,-1,2$. Do you see a pattern?

Answer 2

Let $r_n$ be a residue of $7^n + 11^n\mod 19$. One may note that $$ 7^n + 7^{n+1} + 7^{n+2} = 7^n(1+7+49) = 7^n\cdot3\cdot19 \equiv 0 \mod 19 \\ 11^n + 11^{n+1} + 11^{n+2} = 11^n(1+11+121) = 11^n\cdot 7\cdot19 \equiv 0 \mod 19. $$ Thus $r_n + r_{n+1} + r_{n+2} \equiv 0\mod 19$. This is recurrent formula gives you values for all $r_n$ if you know, for example, values of $r_0$ and $r_1$ (which are easy to calculate): $$ r_0 = 2, \\ r_1 = -1, \\ r_2 = -r_1 - r_2 = -1, \\ r_3 = -r_2 - r_1 = 2, \\ r_4 = -r_3 - r_2 = -1, \\ \dots $$

Answer 3

The residues for $7^n (mod\;19)$ for $n=1..4$ are $7,11,1,7,\dots\;$ and for $11^n (mod\; 19)$ are $11,7,1,11,\dots\;\;$. So the residues of $7^n + 11^n (mod\;19)$ are the sums: $7+11,11+7,1+1,7+11\dots$.