I have two Hamiltonians, $H(p_1, q_1, \dots, p_n, q_n)$ and $H'(p_{n+1}, q_{n+1}, \dots, p_m, q_m)$. They are non-interacting, indeed they have different variables. Moreover, I know that they are both non-integrable. Taken together, the two non-interacting systems are described by the Hamiltonian: $$ H_T(p_1, q_1, \dots, p_m, q_m) = H(p_1, q_1, \dots, p_n, q_n) + H'(p_{n+1}, q_{n+1}, \dots, p_m, q_m) $$
My question is: is it possible that $H_T$ is integrable? I guess "no", but I would like to have a proof.
At first, it could appear as a simple question. Intuitively, "integrable" means "soluble", thus, if the two systems together ($H_T$) are soluble, then a single system alone ($H$) is also soluble.
But "integrable" does not really mean "soluble": it means that it has $m$ constants of motion. So the question is not that simple: it is still possible that $H_T$ has $m$ constants of motion, but a constant of motion of $H_T$ could be not constant for $H$.
More formal statement of the problem
This is a more rigorous formulation of the theorem that I want to prove. Please notice that it is expressed as the contrapositive of what I wrote above.
Two Hamiltonians are given, $H(p_1, q_1, \dots, p_n, q_n)$ and $H'(p_{n+1}, q_{n+1}, \dots, p_m, q_m)$. The Hamiltonian $H_T$ is defined as: $$ H_T(p_1, q_1, \dots, p_m, q_m) = H(p_1, q_1, \dots, p_n, q_n) + H'(p_{n+1}, q_{n+1}, \dots, p_m, q_m) $$
Proposition to be proved: if $H_T$ has at least $n+m$ independent constants of motion, of which $n+m$ in involution, then $H$ has at least $n$ independent constants of motion, of which $n$ in involution.
I would be happy as well to prove much less: if $H_T$ is completely superintegrable (it has $2(n+m)-1$ independent constants of motion, of which $n+m$ in involution), then $H$ has a constant of motion besides $H$ itself. Belive me, also this is tricky.
Important additional information
It is worth noting that, in general, a constant of motion of $H_T$ is not a constant of motion of $H$. It is easy to show an example. Take the equations of a forced (nonlinear) pendulum. It can be written as an autonomous system by adding a degree of freedom: $$ H(p_1, q_1, p_2, q_2) = \frac{p_1^2}{2} + \cos(q_1) + \epsilon q_1 \sin(\omega q_2) + p_2 $$ Clearly, $q_2$ plays the role of time, $q_2=t$. Now, take the same form for $H'$: $$ H'(p_3, q_3, p_4, q_4) = \frac{p_3^2}{2} + \cos(q_3) + \epsilon q_3 \sin(\omega q_4) + p_4 $$
It is easy to show that $H_T=H+H'$ has three constants of motion: $H$, $H'$ and $f=q_4-q_2$. The first two are evident. For the third: $$ \left\{ f, H_T\right\} = \left\{ q_4, H'\right\} - \left\{ q_2, H\right\}= 1-1 = 0 $$ However, this third constant of motion, $f$, is not a constant of motion separately for $H$ and $H'$: $$ \left\{ f, H\right\} = \left\{ q_4-q_2, H\right\} = -1 $$
We can also notice the following. From the knowledge on the non-autonomous forced pendulum, we know that its autonomous version is not integrable. Thus $H$ and $H'$ are non-integrable; since they have only 2 degrees of freedom, this means that they only have one constant of motion, namely $H$ and $H'$, respectively.
Concluding: the answers to this question cannot use the property that a constant of motion of $H_T$ is also a constant of motion of $H$, because this property does not hold in general.
