Canonical transformation problem (Apologies if HW questions are not allowed -- I couldn't really find a definite answer on this)
Question

Let $Q^1 = (q^1)^2, Q^2 = q^1+q^2, P_{\alpha} = P_{\alpha}\left(q,p \right), \alpha = 1,2$ be a CT in two freedoms.
(a) Complete the transformation by finding the most general expression for the $P_{\alpha}$.
(b) Find a particular choice for the $P_{\alpha}$ that will reduce the Hamiltonian
$$H = \left( \frac{p_1 - p_2}{2q^1} \right)^2 + p_2 + (q^1 + q^2)^2$$
to
$$K = P_1^2 + P_2.$$

Attempt
I have shown that
$$P_1 = \frac{1}{2q^1} \left( p_1 + \frac{\partial F}{\partial q^1} - p_2 - \frac{\partial F}{\partial q^2} \right), $$
$$P_2 = p_2 + \frac{\partial F}{\partial q^2}$$
is the most general canonical transformation for the momenta, where $F=F(q^1, q^2)$. This is consistent with the solution manual. For part b, however, the answer I get for an intermediate step is inconsistent with the solutions manual, and I don't understand why. Given that the transformation is canonical, all I need to do to find the transformed Hamiltonian K is find the inverse transformation and plug it in to the Hamiltonian H. The inverse transformation is
$$p_2 = P_2 - \frac{\partial F}{\partial q^2},$$
$$p_1 = 2q^1P_1 + P_2 - \frac{\partial F}{\partial q^1}.$$
Plugging this into H, and renaming H to K since it's in terms of the transformed coordinates we have
$$K = P_1^2 + P_2 - \frac{\partial F}{\partial q^2} + (q^1 + q^2)^2.$$
Since we want K to be
$$K = P_1^2 + P_2,$$
this means
$$\frac{\partial F}{\partial q^2} = (q^1+q^2)^2 = (q^1)^2+(q^2)^2+2q^1q^2.$$
$$F=q^2(q^1)^2 + \frac{1}{3}(q^2)^3 +q^1(q^2)^2 + C.$$
Plugging this into the general transformation I derived I find that
$$P_1 = \frac{1}{2q^1} \left(p_1-p_2-(q^1)^2 \right),$$
$$P_2 = (q^1+q^2)^2+p_2.$$
My equation for $P_2$ is consistent with the solutions manual, but my equation for $P_1$ is not. According to the solutions manual
$$P_1=\frac{p_1+p_2}{2q^1}.$$
My question is, is my methodology essentially correct, and if so did I go wrong in the algebra or did I make some sort of mistake in how I solved the problem.
 A: I think your solution is basically correct.  
Part (a)
To find the missing transformations of the momenta, we first try to find a generating function $\cal F_2(q, P)$ that generates the known transformations of the coordinates.  Then, we use this generating function $\cal F_2(q, P)$ to compute the relations regarding the momenta.
The transformation of coordinates $Q^i = Q^i(q)$
can be conveniently generated by the generating function of type 2 as
\begin{align}
\cal F_2(q, P)
&=\sum_i P_i \, Q^i(q) + F(q),
\end{align}
where $F(q)$ is arbitrary function of $q$.
In this way, the requirement
\begin{align}
\frac{ \partial \cal F_2(q, P) }{ \partial P_i} = Q^i(q).
\end{align}
is automatically satisfied.
In our case
\begin{align}
\cal F_2(q, P)
&= P_1 \, Q^1(q^1, q^2)
+ P_2 \, Q^2(q^1, q^2)
- F \\
&= P_1 \, (q^1)^2
+ P_2 \, (q^1 + q^2)
- F,
\end{align}
where $F \equiv F(q^1, q^2)$ is an arbitrary function of $q^1$ and $q^2$.
So
\begin{align}
p_1 &= \frac{ \partial \cal F_2(q, P) }{ \partial q^1 } = 2 P_1 \, q^1 + P_2
 - \frac{\partial F }{\partial q^1}, \\
p_2 &= \frac{ \partial \cal F_2(q, P) }{ \partial q^2 }
 = P_2 - \frac{\partial F }{\partial q^2}.
\end{align}
Or
\begin{align}
P_1 &= \frac{1}{2q^1} \left(
p_1 + \frac{ \partial F } { \partial q^1 }
-p_2 - \frac{ \partial F } { \partial q^2 }
\right)
\tag{1}
\\
P_2 &= p_2 + \frac{ \partial F } { \partial q^2 }.
\tag{2}
\end{align}
Part (b)
Basically we need to find an $F$ such that $K$ matches $H$, because
$$
d{\cal F}_2 = p \, dq + Q dP + (K - H) \, dt,
$$
and our $F_2$ does not depend on time explicitly (so $K-H$ must vanish).
Now by the solution of part (a), we have
\begin{align}
H &= \left( \frac{p_1 - p_2}{2q^1} \right)^2 + p_2 + (q^1 + q^2)^2, \\
K &= P_1^2 + P_2 \\
 &= \left( \frac{p_1 - p_2 + \partial F/\partial q^1 - \partial F/\partial q^2}{2q^1} \right)^2 + p_2 + \partial F / \partial q^2.
\end{align}
It would be nice if
\begin{align}
\partial F/\partial q^1 &= \partial F/\partial q^2,
\\
\partial F/\partial q^2 &= (q^1 + q^2)^2.
\end{align}
A simple solution would be 
\begin{align}
F = \frac{1}{3} (q^1 + q^2)^3.
\end{align}
Then Eq. (1) and (2) means
\begin{align}
P_1 &= \frac{1}{2q^1} \left(
p_1 - p_2
\right)
\tag{1}
\\
P_2 &= p_2 + (q^1 + q^2)^2.
\tag{2}
\end{align}
The result $P_1 = (p_1 + p_2)/(2q^1)$ doesn't make sense, because it implies $\dot P_1 \ne 0 = -\partial K/\partial Q^1$.
So the plus sign might be a typo.
