# If $a\in R$ and the equation $-3(x-\lfloor x \rfloor)^2+2(x-\lfloor x \rfloor)+a^2=0$ has no integral solution,then all possible values of $a$

If $a\in R$ and the equation $-3(x-\lfloor x \rfloor)^2+2(x-\lfloor x \rfloor)+a^2=0$ has no integral solution, then all possible values of $a$ lie in the interval

$(A)(-1,0)\cup(0,1)$
$(B)(1,2)$
$(C)(-2,-1)$
$(D)(-\infty,-2)\cup(2,\infty)$

My try:
Let $x-\lfloor x \rfloor = \{x\}= t$, where $0 \leq \{x\}<1\Rightarrow 0\leq t<1$. Then

$$\Rightarrow -3t^2+2t+a^2 = 0\Rightarrow 3t^2-2t-a^2 = 0$$

$$\displaystyle \Rightarrow t = \frac{2\pm \sqrt{4+12a^2}}{6} = \frac{1\pm \sqrt{1+3a^2}}{3}$$

I do not know how to solve further.

• Integral solution means $t=0$. There is one only if $a=0$, so possible values of $a$ are $\mathbb{R} \setminus \{0\}$. Are you sure there are no more restrictions (like $-3y^2+2y+a^2$ having a solution)? – Abstraction Mar 10 '16 at 9:21
• – Martin Sleziak Aug 12 '17 at 7:52

EDIT (ELABORATION)

There is probably a typo here; it should say real solutions.

Assume it has solutions. $$0\leq t<1, t=\frac{1\pm \sqrt{1+3a^2}}{3} \Leftrightarrow 1 \le \sqrt{1+3a^2} < 2$$ From the fact that $1-\sqrt{1+3a^2}<0$ if $a \neq 0$. From here, we square both sides to get $$0 \le a^2 \le 1$$ This implies $-1 \le a \le 1$.

• Yes i am getting answer with $1\leq\sqrt{1+3a^2}<2$ but i do not understand how you got this one.@MXYMXY – Brahmagupta Mar 10 '16 at 9:32
• @Brahmagupta The smaller value among $\frac{1\pm \sqrt{1+3a^2}}{3}$ would be $\frac{1-\sqrt{1+3a^2}}{3}$, the larger value would be $\frac{1+ \sqrt{1+3a^2}}{3}$ The minimum value would be less or equal to $0$, and the maximum value would be less than $1$ – S.C.B. Mar 10 '16 at 9:35

First, let us see for which values of $a$ there will be solutions:

Since $\,0\leq t = \left\lfloor x \right\rfloor<1\,$ we have

\begin{align} t &= \dfrac{1\pm \sqrt{1+3a^2}}{3} \in \left[0,1\right) &\implies&& 0\le1\pm\sqrt{1+3a^2} < 3 \\ &\big(\text{subtract } 1 \text{ from each inequality}\big) &\implies&& -1\le\pm\sqrt{1+3a^2} < 2 \end{align} Now consider two cases separately:

1. $\,-1\le + \sqrt{1+3a^2} < 2\,$ \begin{align} &-1\le \sqrt{1+3a^2} < 2&\implies&& 0\le\sqrt{1+3a^2} < 2 \\ &\big(\text{square both sides of each inequality}\big) &\implies&& 0\leq 1+3a^2 <4 \\ &\big(\text{subtract } 1\big) &\implies&& 0\leq 3a^2<3 \\ &\big(\text{divide by } 3\big) &\implies&& 0\leq a^2 < 1 \\ &&\implies&& \bbox[1ex, border:solid 1.5pt #e10000]{a\in\left(-1,1\right)} \end{align}
2. $\,-1\le - \sqrt{1+3a^2} < 2\,$ \begin{align} &-1\le- \sqrt{1+3a^2} < 2&\implies&& 0 < \sqrt{1+3a^2} \leq 1 \\ &\big(\text{square both sides of each inequality}\big) &\implies&& 0\lt 1+3a^2 \leq 1 \\ &\big(\text{subtract } 1 \text{ again}\big) &\implies&& -1\lt 3a^2 \leq 0 \\ &&\implies&& \bbox[1ex, border:solid 1.5pt #e10000]{a = 0} \end{align}

Second, find which values of $a$ correspond to the integral solution of the equation. The answer will be complement to the set of these values in $\mathbb R$.