Product-like structures on spheres For $i=1,2$, let $j_i$ denote the inclusion of $S^n$ into the product $S^n \times S^n$ as the $i^{\text{th}}$ factor. I would very much like to know the answer to the following question, which seems to be very basic: 
For which $(a_1,a_2) \in \mathbb Z^2$ is there a map $$m \colon S^n \times S^n \to S^n,$$ such that the map $$m \circ j_i\colon S^n \to S^n$$ has degree $a_i$?
Here are some thoughts: 
(1) If we can do it for $(a_1,a_2)$ then we can also do it for $(\lambda_1a_1,\lambda_2a_2)$ for any $\lambda_1,\lambda_2$, by precomposing with a suitable map.
(2) If $n$ is even, there does not exist such a map if both $a_1$ and $a_2$ are nonzero: Indeed, if $m$ was as required, the induced map $H^{\ast}(S^n \times S^n) \leftarrow H^{\ast}(S^n)$ sends the fundamental class $[S^n] \in H^n(S^n)$ to $a_1([S^n] \times 1) + a_2 (1 \times [S^n])$. But then $[S^n] \cup [S^n] = 0$ gets send to (here we use that $n$ is even) $a_1a_2([S^n] \times [S^n]) \neq 0$, a contradiction.
(3) If $n = 1,3$ or $7$, then $S^n$ is an $H$-space, which means we can find a $m$ for $(1,1)$ and hence for all $(a_1,a_2)$.
(4) If $n$ is odd but different from $1,3,7$, it is known (by Adams) that $S^n$ is not an H-space, hence we can not find an $m$ for $(1,1)$. Can we find one for $(2,2)$ or even $(1,2)$?
(5) The question can be rephrased in terms of the homotopy group $\pi_{2n-1}(S^n \vee S^n)$ which can be computed using Hilton's theorem, but I do not see how this can be useful.
 A: Your condition determines the map $a_1 \vee a_2 : S^n \vee S^n \to S^n$ on the $n$-skeleton, so the question is when does this extend over the $2n$-cell of $S^n \times S^n$. The $2n$-cell is by definition attached along the universal Whitehead product $[\iota_1, \iota_2 ] \in \pi_{2n-1}(S^n \vee S^n)$, so the question becomes under what conditions is
$$[a_1, a_2] =0 \in \pi_{2n-1}(S^n).$$
The Whitehead product is bilinear, so this asks when $a_1 a_2 [\iota, \iota] =0 \in \pi_{2n-1}(S^n)$. 
As you point out, when $n$ is even $[\iota, \iota]$ has infinite order (as it has Hopf invariant $\pm 2$), so this happens if and only if $a_1 a_2=0$.
If $n$ is odd then the graded-commutativity of the Whitehead product means that $2[\iota,\iota]=0$. Now it follows from the EHP sequence that $[\iota, \iota]=0$ if and only if $\pi_{2n+1}(S^{n+1})$ has an element of Hopf invariant 1, which by Adams means $n=1,3,7$. Thus if $n=1,3,7$ then there are no conditions on the $a_i$, and otherwise the map exists if and only if $a_1 a_2$ is even.
