Let $p=p(x,y),q=q(x,y) \in \mathbb{C}[x,y]$ be a Jacobian pair, namely, $p_xq_y-p_yq_x \in \mathbb{C}^*$. Denote $a:= \deg(p)$ and $b:= \deg(q)$, where $\deg()$ denotes the total degree ($(1,1)$-degree).

There are several nice results about when a Jacobian pair is an automorphic pair (= $f: (x,y) \mapsto (p,q)$ is an automorphism of $\mathbb{C}[x,y]$) involving $a$ and $b$.

Each of the following conditions implies that $f$ is an automorphism:

**(1)** $\gcd(a,b)=1$; Magnus.

**(2)** $\gcd(a,b) \leq 2$; Nakai-Baba

**(3)** $\gcd(a,b)=P$, $P$ is a prime number; Appelgate-Onishi-Nagata.

**(4)** $a=PQ$, $\{P,Q\}$ are prime numbers; Nagata.
See van den Essen's book, pages 254-255.

Is there any progress on the following two natural generalizations:

(3)'$\gcd(a,b)=PQ$, $\{P,Q\}$ are prime numbers.

(4)'$a=PQR$, $\{P,Q,R\}$ are prime numbers.I guess it is quite difficult to prove $(3)'$ or $(4)'$, so let us concentrate on the special case where $P=2$ and $Q \geq 5$, call it $(3)''$: $\gcd(a,b)=2Q$, $Q$ is a prime number $\geq 5$ (for $\gcd(a,b) \in \{4,6\}$ it is already known that $f$ is an automorphism).

Unfortunately, $(3)''$ does not imply $(4)''$: $a=2QR$, $\{Q,R\}$ are prime numbers, since it may happen that $\gcd(a,b)=QR$ with $Q \neq 2, R \neq 2$ (all the other options for $\gcd(a,b)$ are ok: $\{1,2,Q,R,2Q,2R\}$, in case we have $(3)''$).

**Remarks:**

**(I)** It is easy to prove that $(3)'$ implies $(4)'$, by similar arguments to those that show that $(3)$ implies $(4)$, which can be found in the above mentioned Nagata's paper on the first lines on page 170.

**(II)** I can prove that $(4)'$ implies $(3)'$ (as well as that $(4)$ implies $(3)$).

**(III)** Probably it is possible to replace $\mathbb{C}$ by any algebraically closed field of characteristic zero or even by any field of characteristic zero, but I do not mind to work over $\mathbb{C}$.

**(IV)** It seems that the difficult step is to move from $\gcd(a,b)=P$ to $\gcd(a,b)=PQ$ or even just to $\gcd(a,b)=2P$, but we are optimistic (at least for $\gcd(a,b)=2P$).

Any hints and comments are welcome!

Edit:I have recently found this paper, in which $(3)''$ is proved; I wonder whether or not it is easy to first generalize $\gcd(a,b)=2Q$ to $\gcd(a,b)=3Q$ and then to $\gcd(a,b)=PQ$. See also this paper, which contains the same result.