# 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

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Note that $-a$ is the sum and $b$ is the product of the roots. If the roots are integer, their sum is odd only if they have different parity, their product is odd only if they both are odd. Hence no solution.

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awesome observation man ! simple but very good solution ! – lsp Jun 7 '13 at 8:58
very nice observation – trying Jun 7 '13 at 9:56

equation has solution $\iff$ $y\in \mathbb Z \iff \displaystyle \frac{-a\pm \sqrt{a^2-4b}}{2}\in \mathbb Z \iff -a\pm \sqrt{a^2-4b}=2k :k\in \mathbb Z$ $$\iff (a+2k)^2=a^2 - 4b\iff a^2-(a+2k)^2=4b \iff$$ $$(-2k)(2a+2k)=4b\iff k(a+k)=b$$ if $k=2m$ and $a$ be odd then $b$ is even and if $k=2m+1$ and $a$ be odd then $b$ is even. Similarly, we have for $b$. Finally we conclude equation has solution when $a$ or $b$ be odd.

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Babak S.: thanks and congratulation happy eid – Maisam Hedyelloo Jun 7 '13 at 9:37
Happy one to you Maisam. El-Thomas-2A – Babak S. Jun 7 '13 at 9:46

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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