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A cute problem

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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. :)

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2 thoughts on “A cute problem”

  1. 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

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