solving the congruence: $5^n\equiv3^n+2 \pmod{11}.$ I'd really like your help with solving the following congruence:
$$5^n\equiv3^n+2 \pmod{11}.$$
I don't know with what to start.
Any help? 
Thanks
 A: Just note that 11 is prime so you only need to consider n modulo 10 here. Just list $5^n$ and $3^n$ modulo 11 for n going from 0 to 9:
$n\,\;\quad 0\quad  1\quad  2\quad  3\quad  4\quad  5\quad  6\quad  7\quad  8\quad  9$
$5^n\quad 1\quad 5\quad  3\quad  4\quad  9\quad  1\quad  5\quad  3\quad  4\quad  9$
$3^n\quad  1\quad  3\quad  9\quad  5\quad  4\quad  1\quad  3\quad  9\quad  5\quad  4$
Easy to see that this only happens at $n$ being $1$ or $6$ modulo $10$. So in other words we want $n$ to be $1$ modulo $5$.
A: We observe that $\mathrm{ord}_{11}(3) = \mathrm{ord}_{11}(5)=5$
So, the minimal non-negative solution(s) if any will lie between 0 and 4.
Clearly, $s=1$.
Now $5^{5t+s}-3^{5t+s} \equiv 5^s-3^s \mod\ 11$
So, the general solution will be $5t+1$ where $t$ is any integer.
A: Brute force works quickly as others have noted. Here is another way: mod $\,11,\,\ 3 \equiv 5^2,\,$ hence $\rm\, 0 \,\equiv\, 3^n\!-\!5^n\!+\! 2\, \equiv\, 5^{2n}\!-\!5^n\!+\!2\,\equiv\, x^2-x+2\,\equiv\, (x+\color{#C00}4)(x-\color{#0A0}5),\ $ for $\rm\, x = 5^n.\,$ Noticing that the sequence $\rm\, 5^n \equiv 1,\,5,\,3,\,4,\,-2,\,1,5,\,\ldots$ $\Rightarrow$ $\rm\,5^n = x\not\equiv -\color{#C00}4,\,$ and $\rm\,\color{#0A0}5\equiv x\!\iff\! n\equiv 1\pmod 5.$
A: For $n=0,1,2,\cdots,10 $, $5^n-3^n=0,2,5,10,5,0,2,5,10,5,0\implies $ only solutions for $5^n-3^n\equiv 2\pmod {11}$ are $n\equiv1\pmod {5}$(as the pattern repeats itself after $5$ terms)
