House Function

Let $a = n + r$ where $n \in \mathbb{N}$ and $0 < r < 1$ . Then

$\lfloor a \rfloor \lceil a \rceil = a^2 \Rightarrow n (n + 1) = (n + r)^2 \Rightarrow r^2 + 2 n r - n = 0$

Solving the quadratic equation with respect to $r$ yields

$r = -n \pm \sqrt{n (n + 1)}$

Since $n^2 < n (n + 1) < (n + 1)^2$ , we see that $0 < -n + \sqrt{n (n + 1)} < 1$ and thus,

$a = n + r = n + (-n + \sqrt{n (n + 1)}) = \sqrt{n (n + 1)}$

Hence the three least, non-integral, positive solutions to $\lfloor a \rfloor \lceil a \rceil = a^2$ are found when $n = 1, 2, 3$ , that is $\sqrt{2}, \sqrt{6}, 2\sqrt{3}$ .

The product of the solutions is

$2 \sqrt{2} \sqrt{3} \sqrt{6} = 2 \sqrt{6} \sqrt{6} = 12$

If a was a fraction, it's obvious that $a^2$ is non integral, but the LHS is guarnteed integral. Thus a is a square root of some number.

We must check for the first few numbers hat can be represented as $n(n+1)$ . Thus plugging one, two, and three into this, we get $\sqrt2, \sqrt6, \sqrt{12}$

............................................................................................................................................ $a=n+r.~ Hence ~ n^2 + 2 n * r + r^2 = a^2 =n(n + 1).$ $Hence~ integer~ n= \dfrac{r^2}{1 - 2r}$

On an IT calculator. Draw $y= \dfrac{r^2}{1 - 2r}$ graph and y=1, 2, 3..

It intersects 1 at 0.41421356, 2 at 0.44948974, 3 at 0.46410162.

1.41421356 * 2.44948974 * 3.46410162 = 12.

Denote A=a+x is a number that satisfies the equation (0<=x<1): $a(a+1)=(a+x)^2$ $a^2+a=a^2+2ax+x^2$ $x^2+2ax-a=0$ $x=\sqrt{a^2+4a}-a$ Let a=1,2,3 We get x=0.414214, 0.44949, 0.4641. $Ans = \boxed{12}$