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- Feb 14, 2012

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- #1

- Feb 14, 2012

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- Feb 14, 2012

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$\lfloor x \rfloor \le x < \lfloor x \rfloor +1$

Then putting $\lfloor x \rfloor =y$ and $x-\lfloor x \rfloor=z$ we have

$z^2+z-y^2+y-k=0$, where $y$ is an integer and $z\in [0,\,1)$.

Expressing $z$ in terms of $y$ yields

$z=\dfrac{-1\pm\sqrt{1+4(y^2-y+k}}{2}$.

Since $z\ge 0$, we have

$z=\dfrac{-1+\sqrt{1+4(y^2-y+k)}}{2}$

So $0\le \dfrac{-1+\sqrt{1+4(y^2-y+k)}}{2}<1$, or equivalently,

$0\le y^2-y+k<2$

If $x_1>x_2$ are two distinct non-negative roots of the given equation, then $y_1>y_2$. Indeed, since $\lfloor x_i \rfloor=y_i$ and $x_i-\lfloor x_i \rfloor =z_i$ $(i=1,\,2)$, we have $y_1\ge y_2$. Assume that $y_1=y_2$. In this case, by $z=\dfrac{-1+\sqrt{1+4(y^2-y+k}}{2}$, $z_1=z_2$ and so $x_1=x_2$. This is impossible.

Thus $y_1>y_2$. From $0\le y^2-y+k<2$, it follows that

$|y_1^2-y_1-y_2^2+y_2|<2$, or equivalently,

$(y_1-y_2)|y_1+y_2-1|<2$.

Note that $y_1,\,y_2$ are integers, and so $y_1-y_2\ge 1$. Then the last inequality shows that $|y_1+y_2-1|=0,\,1$.

For $|y_1+y_2-1|=0$: $y_1+y_2=1$ and hence $y_1=1,\,y_2=0$.

For $|y_1+y_2-1|=1$: $y_1+y_2=2$ and hence $y_1=2,\,y_2=0$. But these values do not satisfy $(y_1-y_2)|y_1+y_2-1|<2$.

Thus we see that if the given equation has two non-negative distinct roots $x_1>x_2$, then $\lfloor x_1 \rfloor=1,\,\lfloor x_2 \rfloor=0$. Hence,

$x_1=\dfrac{\sqrt{1+4k}+1}{2},\\x_2=\dfrac{\sqrt{1+4k}-1}{2}$

Obviously, this equation cannot have more than two distinct roots.

Finally, it follows that the possible range of $k$ is $0\le k <2$.

- Aug 30, 2012

- 1,157

Nice!

-Dan

-Dan