<s>[INST] no. of solution of the equation $[x]^2+a[x]+b = 0$ is If $a$ and $b$ are odd integer. Then the no. of solution of the  equation $[x]^2+a[x]+b = 0$ is 
where $[x] = $ greatest Integer function
My Try:: Let $[x] = y$. Then equation become $y^2+ay+b = 0$
Now If given equation has real Roots, Then $\displaystyle y = \frac{-a\pm \sqrt{a^2-4b}}{2}$
Now $a^2-4b = k^2\Leftrightarrow a^2-k^2=4b^2$. where $k\in \mathbb{Z}$
Now How can I solve after that.
Help required
Thanks
 [/INST] If $y := \lfloor x \rfloor$ is even, $y^2$ is even, $ay$ is even, $b$ is odd, hence $y^2 + ay + b$ is odd, so $y^2 + ay+b \ne 0$. If on the other hand, $y$ is odd, $y^2$ and $ay$ are odd also, as is $b$, so $y^2 + ay+ b$ is odd again. Hence $y^2 + ay  + b$ is an odd integer for every integer $y$. So the equation does not have any solution. 
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