The closed form solution is $P_j = 0, j \ge 0$

I shall show now that there is no solution to the recursive equations with 

(1)  $P_j \gt 0, j = 1, 2, 3, ...$ 

First from 

$\alpha P_1 = P_2 + P_3$

we conclude 

$\alpha \gt 0$

Notice also the $P_0$ appears only in the relation

$\alpha P_0 = P_1$    

which shows that 

$P_0 \gt 0$ as well, but $P_0$ does not appear in the normalization.
Therefore we consider it as a mere abrevíation for $P_1/ \alpha $.

Now we transform the recursive relation into a standard form, which we define here to be one in which an element with a specific index is defined in terms of elements with smaller indices.

Define

(2) $Q_i = P_{i+1}+P_{i+2}+..., i = 0,1,2,...$

As a sum over positive quantities we have 

$Q_i \gt 0, i = 0, 1, 2, ...$

The inversion of (2) is

(3) $P_i = Q_{i-1} - Q_i	, i = 1, 2, ...	$

Now the equations become

$\alpha P_j = Q_j - Q_{2j+1}, j =1, 2, 3, ...    $

Using (3) we get

$\alpha (Q_{j-1}-Q_j) = Q_j - Q_{2j+1}$

or

q[2j+1] = (1+a) q[j] - a q[j-1], j = 1, 2, ...	(4)

This is now a recursive relation in standard form.

The inital values are

q[0] = p[1] + p[2] + ... = 1

because of the normalization condition.

And

q[1] = 1 - p[1] = 1 - a p[0]

can be considered as a free parameter in the interval (0,1).

Before we solve (4) we observe that it defines only the elements with an odd index.
Therefore we let generally

q[2k] = c[k] > 0, k = 1, 2, ...

with positive c[i].

Performing now the first few steps of the solution to (4) we find that

p[10]  = - c[5] - a (1+a) q[1]

But this is a negative quantity, and the contradiction proves the statement.