How can we simplify the expression $P+\sqrt{P^2+\sqrt{P^4+\sqrt{ P^8+\cdots)}}}$? Is there a way to reduce the expression
$P+\sqrt{P^2+\sqrt{P^4+\sqrt{ P^8+\cdots)}}}$?
 A: So our target expression is:
$$ P + \sqrt{P^2 + \sqrt{P^4...}}$$
An indirect approach that is quite simple is to consider
$$ P \left( 1 + \sqrt{1 + \sqrt{1 ...}} \right)$$
It follows that of we distribute the P we will have 
$$ P +P \sqrt{1 + \sqrt{1 ...}} $$
Which yields
$$ P + \sqrt{P^2 + P^2\sqrt{1 + ...}}$$
And again we can distribute under the next root for $P^4$ then $P^8$ etc...
So now the trick is to evaluate:
$$S = 1+\sqrt{1+\sqrt{...}}$$
It follows that 
$$(S-1)^2 = 1+\sqrt{1+\sqrt{...}}$$
And therefore 
$$ (S-1)^2 = S$$
Using this we can solve and find
$$S^2-3S + 1 = 0$$
Which has solutions given by the quadratic formula:
$$S = \frac{3 \pm \sqrt{5}}{2}$$
A simple test of evaluating the first few terms of 
$$S = 1+\sqrt{1+\sqrt{...}}$$
Suggests that solution closer to $2$ is in order this we have 
$$S =  \frac{3 + \sqrt{5}}{2}$$
And this to evaluate:
$$ P + \sqrt{P^2 + P^2\sqrt{1 + ...}}$$
Is just
$$P \left(   \frac{3 + \sqrt{5}}{2}   \right)$$
A: If the series converges to $S$ then $S= P + \sqrt{P^2+\sqrt{P^4+\cdots}} = P\left(1+\sqrt{1+\sqrt{1+\cdots}}\right) = PA$, 
where $A$ is $1 + \sqrt{1+\sqrt{1+\cdots}}$. 
Now, $(A-1)^{2}=A$ so  
$A$= $\frac{3+\sqrt 5}{2}$ or $\frac{3-\sqrt 5}{2}$, but we discard the latter root, because if you look at the series $A$ it is obvious that $A>2$.
Therefore,  $S=\frac{3+\sqrt 5}{2}P$.
