Here is a problem that popped up in my head about a year ago.
Problem 1. A function is polynomial of each argument, i.e. for every the function can be represented by a polynomial with integer coefficients, and the same goes for for every . Is it true then that must be polynomial of both arguments simultaneously, i.e. representable by an element of ?
If you’ve ever been a freshman in an analysis course, you are bound to be aware of a common trap with the notion of differentiability. The trap is like this: you are asked to prove that a function is differentiable everywhere. You have several months of single-variable analysis behind your back. So your first instinct may be to prove that and exist at every point and be done with it. But this is not enough, as demonstrated by
Problem 1 asks if there is the same trap with polynomiality (not a word?) as there is with differentiability. When I came up with problem 1 I was busy with more “serious” math, so the question didn’t get any real thought. I always suspected that it would’t take long to solve, but somehow I was too lazy to do the job. That is, I was until this morning, when I finally put in the required several minutes and got this thing sorted out. If you’re curious, you might want to do the same before moving on to the next paragraph.
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