Proof for the Ramanujan's formula ?

$x+n+a=\sqrt{ax+(n+a)^2+x \sqrt{a(x+n)+(n+a)^2 +(x+n) \sqrt\cdots}}$

Proof :

$F(x)=\sqrt{ax+(n+a)^2+x \sqrt{a(x+n)+(n+a)^2 +(x+n) \sqrt{a(x+2n)+(n+a)^2+(x+2n) \sqrt{\cdots}}}}$

 $\Rightarrow F(x)^2=ax+(n+a)^2+x \sqrt{a(x+n)+(n+a)^2 +(x+n) \sqrt{\cdots}}$

$\Rightarrow F(x)^2=ax+(n+a)^2+xF(x+n)\tag{1}\label{1}$  

Let's assume that $F(x)=mx+k$

$\deg F(x)$  cannot be more than  $1$  otherwise left side degree will be bigger than right side in Equation (\ref{1})


$ F(x)^2=ax+(n+a)^2+xF(x+n)$
$\Rightarrow (mx+k)^2=ax+(n+a)^2+x(mx+k)$
$\Rightarrow  m^2x^2+2mkx+k^2=ax+(n+a)^2+mx^2+(mn+k)x$

$m^2=m$
$m=1$  or $m=0$
$2mk=mn+k+a$
If $m=1$, then
$2k=n+a+k$
$k=n+a$

And It is also confirming constant term of
 $m^2x^2+2mkx+k^2=ax+(n+a)^2+mx^2+(mn+k)x$
$\vdots$
$k^2=(n+a)^2$
$F(x)=mx+k$
$F(x)=x+n+a$

$x+n+a=\sqrt{ax+(n+a)^2+x \sqrt{a(x+n)+(n+a)^2 +(x+n) \sqrt{\cdots}}}$

Setting $a=0, n=1$,

$x+1=\sqrt{1+x \sqrt{1 +(x+1) \sqrt{\cdots }}}$