Let's play with sums!

For $x=x_{1}$ and $x=x_{2}$

$\begin{aligned} x^{2}-6x+1=x^{2}-x+1&=0 \text{ (mod 5)} \\ (x^{2}-x+1)(x+1)&=x^{3}+1=0 \text{ (mod 5)} \\ x^{3}&=-1 \text{ (mod 5)} \implies x_{1}^{3}=x_{2}^{3}=-1 \text { (mod 5)}\end{aligned}$

And by vieta, we have $x_{1}+x_{2}=6$ and $x_{1}x_{2}=1$

Then, $S_{50} \text{ (mod 5)}$ is $S_{50}= (x_{1}^{50}+x_{2}^{50}) = ((x_{1}^{3})^{16} \times x_{1}^2 +(x_{2}^{3})^{16} \times x_{2}^2) \text{ (mod 5)} \\ S_{50} = ((-1)^{16} \times x_{1}^2 +(-1)^{16} \times x_{2}^2) \text{ (mod 5)} \\ S_{50} = (x_{1}^2 +x_{2}^2) =((x_{1}+x_{2})^2-2x_{1}x_{2}) \text{ (mod 5)} \\ S_{50} = ((6)^2-2 \times 1) \text{ (mod 5)} \\ S_{50} = 34 \text{ (mod 5)} \\ S_{50} = \boxed{4} \text{ (mod 5)}$

Since $x_1$ and $x_2$ are roots of $x^2-6x+1=0$ , by Vieta's formula we have $x_1+x_2 = 6$ and $x_1x_2 = 1$ . And by Newton sums method, we have:

$\begin{aligned} S_1 & = x_1+x_2 = 6 \\ S_2 & = x_1^2 + x_2^2 = (x_1+x_2)^2 - 2x_1x_2 = 34 \\ S_n & = (x_1+x_2)S_{n-1}-x_1x_2S_{n-2} & \small \color{#3D99F6} \text{for } n \ge 3 \\ & = 6S_{n-1} - S_{n-2} \end{aligned}$

This implies that $S_n$ has a recurrence relation and its characteristic equation is:

$\begin{aligned} r^2 - 6r + 1& = 0 \\ r & = 3 \pm \sqrt 8 \\ \implies S_n & = c_1 (3+\sqrt 8)^n + c_2 (3-\sqrt 8)^n & \small \color{#3D99F6} \text{where }c_1 \text{ and } c_2 \text{ are constants.} \\ S_1 & = (3+\sqrt 8)c_1 + (3-\sqrt 8)c_2 \\ S_2 & = (3+\sqrt 8)^2c_1 + (3-\sqrt 8)^2c_2 \\ & = (17+6\sqrt 8)c_1 + (17-6\sqrt 8)c_2 \end{aligned}$

$\implies \begin{cases} S_1: & 3(c_1+c_2) + \sqrt 8(c_1-c_2) = 6 \\ S_2: & 17(c_1+c_2) + 6\sqrt 8(c_1-c_2) = 34 \end{cases}$

$\begin{aligned} 6S_1 - S_2: \quad c_1+c_2 & = 2 \end{aligned}$

$\begin{aligned} S_1: \quad 3(c_1+c_2) + \sqrt 8(c_1-c_2) & = 6 \\ 3(2) + \sqrt 8(c_1-c_2) & = 6 \\ c_1-c_2 & = 0 \\ \implies c_1 & = c_2 = 1 \end{aligned}$

Therefore,

$\begin{aligned} S_n & = (3+\sqrt 8)^n + (3-\sqrt 8)^n \\ S_{50} & = (3+\sqrt 8)^{50} + (3-\sqrt 8)^{50} \\ S_{50} & \equiv 2 \sum_{k=0}^{25} \binom {50}{2k} 3^{2k}\cdot 8^k \text{ (mod 5)} & \small \color{#3D99F6} \text{Note that } \binom {50}{2k} \text{ is divisible by 5 for }1 \le k \le 24 \\ & = 2\left(3^{50} + 8^{25}\right) \text{ (mod 5)} & \small \color{#3D99F6} \text{Since }\gcd (3.5) =1, \gcd (8,5) =1 \text{, Euler's theorem applies.} \\ & = 2\left(3^{\color{#3D99F6} 50 \text{ mod }\phi (5)} + 8^{\color{#3D99F6} 25 \text{ mod }\phi (5)}\right) \text{ (mod 5)} & \small \color{#3D99F6} \text{Euler's totient function }\phi (5) = 4 \\ & = 2\left(3^{\color{#3D99F6} 50 \text{ mod }4} + 8^{\color{#3D99F6} 25 \text{ mod }4}\right) \text{ (mod 5)} \\ & = 2\left(3^{\color{#3D99F6} 2} + 8^{\color{#3D99F6}1}\right) \text{ (mod 5)} \\ & = 2\left(17\right) \text{ (mod 5)} \\ & = \boxed{4} \text{ (mod 5)} \end{aligned}$

On apply vieta's formula, we have $x_1x_2 =1$ or $x_2 = \dfrac{1}{x_1}$

Since $x_1, x_2$ are roots of the given equation so, they must satisfy the equation

Given equation can be written as

$\Longrightarrow x_1 + \dfrac{1}{x_1} = 6$

$S_1\equiv x_1 + x_2 = 6$

$S_2\equiv x_1^2 + x_2^2 = 34$

$S_3\equiv x_1^3 + x_2^3 = 198$

$S_4\equiv x_1^4 + x_2^4 = 1154$

$S_5\equiv x_1^5 + x_2^5 = 6726$

we notice the last digit starts repeated after a cycle of 4, therefore in order to find remainder of 50th power when divided by 5,

we only need to see last digit of the 2nd term of the cycle, which is 4 and hence on dividing by 5 we get remainder as $\boxed 4$ .

If you are curious: $S_{50} = 189482250299273866835746159841800035874$

When you divide by 5, the remainder is $\boxed {4}$