Find a nonzero polynomial $P(x,y)$ where the coefficients are integers such that $P(\lfloor a \rfloor, \lfloor 2a \rfloor) = 0$ $\forall a \in$ $\mathbb{R}$
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$\begingroup$ Please show your try at the question first. $\endgroup$ – Saad Feb 5 '18 at 11:43
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$\begingroup$ Hint: $\lfloor 2a\rfloor$ is an integer larger or equal to $2a-2$ and smaller or equal to $2a+2$. Substituting $a$ with $\lfloor a\rfloor$ in the latter quantities does bring substantial changes to the observation. $\endgroup$ – user228113 Feb 5 '18 at 11:52
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Hint: $\lfloor a \rfloor = a - f$ where $f$ is the fractional part. Then, there are three possibilities:
- $f < 0.5$, giving $\lfloor 2a \rfloor = 2a - 2f$
- $f = 0.5$, giving $\lfloor 2a \rfloor = 2a$
- $f > 0.5$, giving $\lfloor 2a \rfloor = 2a - 2f + 1$
These are easily verified. But case 2 seems problematic in our problem. Can you eliminate case 2 (hint: what is $2f$ in case 2)?
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$\begingroup$ I would rather discuss the possibilities in terms of the relation between $\lfloor a\rfloor$ and $\lfloor2a\rfloor$. On a side note, have you published an OEIS entry on math.stackexchange.com/questions/2469058/…? $\endgroup$ – Ivan Neretin Feb 5 '18 at 12:14
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$\begingroup$ @IvanNeretin Oops! I have totally forgotten. I'll see if I can get to it later today. $\endgroup$ – orlp Feb 5 '18 at 12:43
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Hint: $\,2 \lfloor a \rfloor -1 \le 2a-1 \lt \lfloor 2a \rfloor \le 2a \lt 2 \left(\lfloor a \rfloor +1\right) = 2 \lfloor a \rfloor + 2\,$, so $2 \lfloor a \rfloor\le \lfloor 2a \rfloor \lt 2 \lfloor a \rfloor + 2\,$. It follows that $\,\lfloor 2a \rfloor\,$ can only take one of the two integer values $\,2 \lfloor a \rfloor\,$ or $\,2 \lfloor a \rfloor+1\,$, and therefore:
$$\big(\lfloor 2a \rfloor - 2 \lfloor a \rfloor \big)\big(\lfloor 2a \rfloor - 2 \lfloor a \rfloor - 1\big) = 0 \quad\quad\forall a \in \mathbb{R}$$
