solving $3(x_{n+1} - x_{n}) = \sqrt{16+x^2_n} +\sqrt{16+x^2_{n+1}}$ with $x_1=3$ I was solving a question which led to the following recurrence:
$$3(x_{n+1} - x_{n}) = \sqrt{16+x^2_n} +\sqrt{16+x^2_{n+1}}$$
which I could not solve
my approach :
I tried putting $x_n= tan(t_n)$ and other trigonometric substitutions but could not reach an answer .Is there a way to solve this ?
all help is greatly appreciated.
EDIT:I just wanted to add one thing that $x_1=3$,I hope it makes the question more clear.
 A: Continuing on hamam_Abdallah's hint we can now use $t=\tanh(\frac u2)$ formulas.
$\dfrac{6t_{n+1}}{1-{t_{n+1}}^2}-\dfrac{6t_{n}}{1-{t_{n}}^2}=\dfrac{1+{t_{n+1}}^2}{1-{t_{n+1}}^2}+\dfrac{1+{t_{n}}^2}{1-{t_{n}}^2}$
which after regrouping and factoring gives $\quad\dfrac{2(t_nt_{n+1}+1)(t_nt_{n+1}+3t_{n+1}-3t_n-1)}{(1-{t_{n+1}}^2)(1-{t_n}^2)}=0$
Since $\tanh=\pm 1$ arises only at infinity, we can simplify by the terms $(1-t^2)$ and
$(1+tt)$.
We are left with $$t_nt_{n+1}+3t_{n+1}-3t_n-1=0$$
From Robert Israel comment, we should have $x_i\to +\infty$ (i.e. $t_i\to 1$)
In order to get rid of the constant let shift the limit towards zero with $z=t-1$
We reach $$z_nz_{n+1}+4z_{n+1}-2z_n=0$$
Since $z_i\neq 0$ (else $x_i=+\infty$) we can rewrite it $\quad 1+\dfrac{4}{z_n}-\dfrac{2}{z_{n+1}}=0\ $ which makes the double product disappear.
So let set $w=\dfrac 2z$ we get: $$w_{n+1}=2w_n+1$$

Finally $\ w_n=\alpha2^n-1$

We can now go back the chain $x=4\sinh(u),\ t=\tanh(\frac u2),\ z=t-1,\ w=\frac 2z$

$$x_n=4\sinh\left(2\tanh^{-1}\left(\frac 2{\alpha 2^n-1}+1\right)\right)$$
I let you calculate $\alpha$ given $x_1=3$, note that it should be negative for $\tanh^{-1}$ to exists.
As suggested in comment, I can replace $x_n=4\sinh(u_n)=\dfrac{8t_n}{1-{t_n}^2}=\cdots=\dfrac 2{\alpha 2^n}-2\alpha 2^n$
$x_1=3=\dfrac 1\alpha-4\alpha\iff \alpha\in\{-1,\frac 14\}$ and since $\alpha<0$ then $\alpha=-1$.
$$\boxed{x_n=2\left(2^n-\frac 1{2^n}\right)}$$
A: hint
put
$$x_n=4\sinh(u_n)$$
then, the relation becomes
$$12(\sinh(u_{n+1})-\sinh(u_n))=4(\cosh(u_n)+\cosh(u_{n+1})$$
A: The simplicity of the result found in my other answer, lead me to think there might be a direct algebraic way to get to a linear recurrence relation for $(x_n)$ with a characteristic equation of degree $2$.
So let's try to introduce $x_{n+2}$ into the game...
For ease of writing, I shall use $\begin{cases}x=x_n\\y=x_{n+1}\\z=x_{n+2}\end{cases}\ $ start with $\ 3(y-x)=\sqrt{16+x^2}+\sqrt{16+y^2}$
Let square each side and put square roots on one side
$$8y^2-18xy+8x^2-32=2\sqrt{16+x^2}\sqrt{16+y^2}$$
Squaring again and factoring gives $$32(2y^2-5xy+2x^2-18)(y-x)^2=0$$
Now as Robert Israel commented, $y-x\ge 8/3\neq 0$ so we are left with the middle term.
Let apply for $n$ and $n+1$ this gives $\begin{cases}2z^2-5yz+2y^2-18=0\\2y^2-5xy+2x^2-18=0\end{cases}$
Subtracting both lines: $2z^2-5yz+5xy-2x^2 = 2(z-x)(z+x)-5y(z-x) = 0$
Again $z-x=(z-y)+(y-x)\ge 2\times \frac 83\neq 0$
Thus we are left with $2(z+x)-5y=0$ which is

$$2x_{n+2}-5x_{n+1}+2x_n=0$$

Characteristic equation is $2r^2-5r+2=0\iff r\in\{2,\frac 12\}$
$$\boxed{x_n=\alpha 2^n+\dfrac{\beta}{2^n}}$$
Solving for initial conditions, lead to the same solution as in my other answer.
