Define an integer sequence by $a_1=4$ and $a_{n+1}=a_n^2$, and another integer sequence by $b_1=3$ and $b_{n+1}=a_n^2-a_nb_n+b_n^2$. It is straightforward to show that $x_n=\frac{a_n}{b_n}$ and that $\sum_{k=1}^{n}x_k ~=~ \frac{(2b_n-a_n)^2}{b_n(a_n-b_n)}$. Since $a_n$ and $b_n$ are relative prime, the numerator of the lowest term expression of $\sum_{k=1}^{n}x_k$ is the perfect square $(2b_n-a_n)^2$.

test20 wrote: Define an integer sequence by $a_1=4$ and $a_{n+1}=a_n^2$, and another integer sequence by $b_1=3$ and $b_{n+1}=a_n^2-a_nb_n+b_n^2$. It is straightforward to show that $x_n=\frac{a_n}{b_n}$ and that $\sum_{k=1}^{n}x_k ~=~ \frac{(2b_n-a_n)^2}{b_n(a_n-b_n)}$. Since $a_n$ and $b_n$ are relative prime, the numerator of the lowest term expression of $\sum_{k=1}^{n}x_k$ is the perfect square $(2b_n-a_n)^2$. How did you end up with $\sum_{k=1}^{n}x_k ~=~ \frac{(2b_n-a_n)^2}{b_n(a_n-b_n)}$? I've tried coming up with it by myself, but I can't seem to simplify it nicely.

Another sollution: Notice that 1/(x(n+1) - 1) = x(n) + x(n-1)… + x(1) + 3 so x(n+1) = (S(n) +4)/(S(n) +3) Hence if S(n) = a/b for some coprime a and b, then S(n+1) = a/b + (a/b + 4)(a/b + 3) = (a+2b)^2/b(3b+a)

Notice that $$x_{n+1}-1=\frac{x_n^2}{x_n^2-x_n+1}-1=\frac{x_n-1}{x_n^2-x_n+1}.$$Therefore, $$\frac{1}{x_{n+1}-1}=\frac{x_n(x_n-1)+1}{x_n-1}=x_n+\frac{1}{x_n-1}\iff x_n=\frac{1}{x_{n+1}-1}-\frac{1}{x_n-1}.$$This means that $$S_k=\sum_{k=1}^nx_k=\frac{1}{x_{n+1}-1}-\frac{1}{x_1-1}=\frac{x_n^2-x_n+1}{x_n-1}-3=\frac{(x_n-2)^2}{x_n-1}.$$ Let $x_n=\frac{p}{q},$ where $\gcd(p,q)=1.$ Hence, after substituting $x_n$ we obtain $$S_k=\frac{(p-2q)^2}{q(p-q)}=\frac{p^2-4pq+4q^2}{q(p-q)}.$$Now since $\gcd(q,p^2-4pq+4q^2)=\gcd(p-q,p^2-4pq+4q^2)=1$, we are done.