A troublesome recurrence

From the first few $x_n$ , it appears that $x_n = 2^n$ . Let us prove the claim by induction that the claim is true for all $n \ge 1$ . We note that the claim is true for $n=1$ and $n=2$ . Assuming the it is true for $n$ . Then

$\begin{aligned} \sum_{k=1}^n \frac {x_{n+1}^2}{x_kx_{k+1}} & = 2 \left(\frac {x_{n+1}^2-x_1^2}{x_2^2 - x_1^2}\right) \\ \frac {x_{n+1}^2}{x_n^2} \sum_{k=1}^{n-1} \frac {x_n^2}{x_kx_{k+1}} + \frac {x_{n+1}}{x_n} & = \frac {x_{n+1}^2 - 4}6 \\ \frac {x_{n+1}^2}{x_n^2} \cdot \frac {x_n^2 - 4}6+ \frac {x_{n+1}}{x_n} & = \frac {x_{n+1}^2 - 4}6 \\ x_{n+1}^2 x_n^2 - 4 x_{n+1}^2 + 6 x_{n+1} x_n & = x_{n+1}^2 x_n^2 - 4 x_n^2 \\ 2 x_{n+1}^2 - 3 x_{n+1} x_n - 2 x_n^2 & = 0 \\ \left(2x_{n+1} + x_n \right) \left(x_{n+1} - 2 x_n \right) & = 0 & \small \blue{\text{Since }x_{n+1} \in \mathbb N} \\ \implies x_{n+1} & = 2x_n = 2^{n+1} \end{aligned}$

Therefore the claim is true for $n+1$ and hence true for all $n \ge 1$ and $x_{10} = 2^{10} = \boxed{1024}$ .

It can be proved inductively that $x_n = 2^n$ :

The first case $n = 2$ is true because $\displaystyle \sum_{i=1}^{2 - 1} \frac{x_2^2}{x_ix_{i + 1}} = \frac{(2^2)^2}{2^1 \cdot 2^2} = 2 = 2 \frac{x_2^2 - x_1^2}{x_2^2 - x_1^2}$ .

Then assuming the sum for $n$ is true, the sum for $n + 1$ is also true because:

$\displaystyle \sum_{i=1}^{(n + 1) - 1} \frac{x_{n + 1}^2}{x_ix_{i + 1}}$

$\displaystyle = \frac{(2^{n + 1})^2}{2^1 \cdot 2^2} + \frac{(2^{n + 1})^2}{2^2 \cdot 2^3} + ... + \frac{(2^{n + 1})^2}{2^n \cdot 2^{n + 1}}$

$\displaystyle = \frac{(2^{n + 1})^2}{2^{2n + 1}}(2^{2n - 2} + 2^{2n - 4} + ... + 1)$

$\displaystyle = 2\bigg(\frac{2^{2n} - 1}{2^2 - 1}\bigg)$

$\displaystyle = \frac{2}{3}(2^{2n} - 1)$

$\displaystyle = \frac{2}{3} \cdot \frac{4(2^{2n} - 1)}{4}$

$\displaystyle = 2 \cdot \frac{(2^{n + 1})^2 - 4}{(2^2)^2 - (2^1)^2}$

$\displaystyle = 2 \cdot \frac{x_{n + 1}^2 - 4}{x_2^2 - x_1^2}$

which completes the inductive proof.

Therefore, $x_{10} = 2^{10} = \boxed{1024}$ .