Quadratic equation with floor function in the variable

Question

Can anybody please explain to me the solution of this problem?

Answer 1

Put t=x-[x] and that gives you $ a^2=3t^2-2t$.now notice that $3t^2-2t$=y can be written as

$$(t-1/3)^2=1/3(y+1/3)$$

Compare this with the equation

$x^2=4ay$

Draw its graph and retain it for x between 0 and 1 because here x--> {x}.when x=1,y=1.so clearly non integral solution arises when $0<a^2<1$. since $a^2>0$.now you should be able to solve it.

Answer 2

Do you not think that $a^2=0$ when $t=\frac 2 3$? So $a\in(-1,1)$?

Seeing that $0<t<1$, where $t=x-[x]$, we need to study the range of $3t^2-2t$ on that domain.

$3t^2-2t$ is a parabola, as shown on the right in the memorandum. The minimum value is at $t=-\frac{-2} 6=\frac 1 3$ which is $\frac 1 3-\frac 2 3=-\frac 1 3<0$. But $a^2$ is assumed nonnegative (i.e. many values of $t$ are disregarded). On the other hand, the two end-values at $t=0$ and $t=1$ are $0$ and $3-2=1$ respectively, so the supremum value of $a^2$ is $1$ and the minimum value is $0$ (because $3t^2-2t$ is continuous; Intermediate Value Theorem).

Now, the second step is to recognize that when $0\leq a^2<1,\quad a\in(-1, 1)$.