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# A cute problem

I came up with this little problem last night. It’s not very difficult to prove but still fun (I think). Here it is : let $A$ be a commutative ring, and let $h(u,v) \in A[u,v]$. Suppose that, for any polynomials $f(u), g(u) \in A[u]$, we have $h(f(u), g(u))=h(g(u), f(u))$. Then $h(u,v)=h(v,u)$.

I’ll post my solution in a couple of days to see if anyone can come up with an alternative solution in the meantime. :)

## 2 thoughts on “A cute problem”

Nice problem! :) Here is my solution:

Suppose that $h(u,v)=\sum_{i,j}a_{i,j}u^iv^j$, then $h(v,u)=\sum_{i,j}a_{j,i}u^iv^j$.

Let $m,n$ be relatively prime integers greater than the degree of $h$. Then the numbers $im+jn$ with $i,j$ not greater than degree of $h$ are all different (exercise!), and so the equality:

$\sum_{i,j}a_{i,j}u^{im+jn}=h(u^m,u^n)=h(u^n,u^m)=\sum_{i,j}a_{j,i}u^{im+jn}$

implies that $a_{i,j}=a_{j,i}$ for all $i,j$ :D

• Perfect solution! It’s exactly the same solution that I found. I wonder if there are any other simple solutions. This is probably as simple as it gets, though.

Congratulations! :D