# 2005 Putnam B1: Find a Polynomial

Find a nonzero polynomial $P(x,y)$ such that $P(\lfloor a\rfloor,\lfloor 2a\rfloor)=0$ for all real numbers $a.$

(Note: $\lfloor v\rfloor$ is the greatest integer less than or equal to $v.$)

I almost get it.

I noticed a pattern. Let $[]$ represent the floor function.

$$[3.2], [6.4] = 3, 6$$

$$\cdots$$

$$[6.4], [12.8] = 6, 12$$

$$\cdots$$

$$[9.9], [19.8] = 9, 19$$

$$2[a] \le [2a]$$

For negative numbers

$$[-1.2], [-2.4] = -2, -3$$

$$[-3.2], [-6.4] = -4, -7$$

$$\cdots$$

$$[-9.9], [-19.8] = -10, -20$$

$$[2a] \le 2[a] + 1$$

This becomes:

$$2[a] \le [2a] \le 2[a] + 1$$

Since, $x = [a]$ and $y = [2a]$ it gives:

$$2x \le y \le 2x + 1$$

What is a possibility now?

For any real number $r$, $\left\lfloor 2r\right\rfloor-2\left\lfloor r\right\rfloor$ can be only zero or one, hence $$P(x,y) = (y-2x)(y-2x-1)$$ works.
• Can you show that: $[2r] - 2[r] = 0, 1$ in cases/ – Amad27 Mar 12 '15 at 15:50
• @Amad27: almost. For sure, $f(x)=\lfloor 2x\rfloor-2\lfloor x\rfloor$ is a $1$-periodic function that can take only integer values, hence you just have to study $f(x)$ on $[0,1]$ to prove my very first claim. – Jack D'Aurizio Mar 12 '15 at 16:50
• @Amad27 The first claim can be seen by your last inequality (in question description), namely $2x\le y\le 2x+1\iff 0\le y-2x\le 1$ (and of course $y-2x$ is an integer). – user26486 Mar 12 '15 at 19:51