Easy Putnam Problem

[Ender]

Find a nonzero polynomial P(x, y) such that P(floor(a), floor(2a)) = 0 for all real a.

floor(a) is defined as the greatest integer less than or equal to a.

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Of course, no googling this problem. Note that an easy putnam problem is not necessarily easy =p. 

EDIT:

Also -- if you post the solution and I don't respond, email it to me, (dijame [[at]] gmail [[dot]] com) and I'll probably get back to you pretty soon.

[rabbit]

Uh.....P(x, y) = x - (y/2) ?

[Ender]

Nope =p

[Ender]

I am going to be AFK for over a week, so if you come up with a solution you can post it here of course, for others to see, but I may not respond.

It's the kind of problem that's trivial to check though. Either plug in some numbers manually, or if you're not satisfied put P(floor(a), floor(2a)) into your calculator with P as a function of a, and it should be 0 for all a.

[Camel]

The problem can be re-written as:

For all real integers i and non-zero polynomial P, P(i,2i)=0 and P(i,2i+1)=0

I think this can be solved with a second degree polynomial by defining the zeros according to the above constraints. When it's not 3AM, I'll give it more though.

[Sidoh]

Here's what I get:

http://t.sidoh.org/?JQUBAQ==

Easy after you realize what Camel pointed out (I got that part independently too).  Putnam problems are usually pretty awesome.

[Ender]

Correct =) (Oh, and I was nonAFK enough to check this every once in a while.)

Remarks:

A polynomial has a finite number of terms. So an answer like P(x, y) = sin(x*pi) - sin(y*pi) wouldn't work, although it would for power series.

The key to this is recognizing that (1) y can only assume values y = 2x or y = 2x + 1 and (2) applying the factored form of polynomials to easily get the solution.

This is one of the ten putnam problems that the most people got correct, but again no putnam problem is easy, as the median is usually like 0.