Adopted from the Vermont Math contest

Let $x=n+s$ where $n$ is an integer and $0\leq s< 1$ , that is $s$ is the fractional part of $x$ .

Substituting this in the original equation we get, $\lfloor x^2 \rfloor-x\lfloor x \rfloor=\lfloor (n+s)^2 \rfloor-(n+s)\lfloor n+s \rfloor=6$ Note that $\lfloor n+s \rfloor=n$ . This implies, $\lfloor n^2+2ns+s^2 \rfloor-(n+s)\ \lfloor n \rfloor=6$ Since $n^2$ is also an integer, $n^2+\lfloor 2ns+s^2 \rfloor- n^2-ns=6\implies\lfloor 2ns+s^2 \rfloor-ns=6$ Observe that $\lfloor 2ns +s^2 \rfloor$ is always an integer (because of the floor function), so for $\lfloor 2ns+s^2 \rfloor-ns$ to be an integer which in this case is $6$ , $ns$ must also be an integer.

Therefore we have, $\lfloor 2ns+s^2 \rfloor-ns=2ns+\lfloor s^2 \rfloor -ns=ns+\lfloor s^2 \rfloor=6$ Since $s<1$ we must have $s^2<1$ , $\implies \lfloor s^2 \rfloor=0$ . Thus the equation becomes, $ns+\lfloor s^2 \rfloor=ns=6\implies s=\frac{6}{n}$

Note that for $s$ to satisfy the condition $0\leq s< 1$ , $n$ must be greater than $6$ $\implies n\geq 7$ . But to achieve the minimum possible value of $x$ , $n$ must be minimized.

$\implies n=7, s= \dfrac{6}{7}\implies x=n+s=7+ \frac{6}{7}= \boxed{\frac{55}{7}}$ .

Therefore, $a=55, b=7 \implies a+b=\boxed{62}$ .

Consider the equation $\lfloor x^2 \rfloor - x \lfloor x \rfloor = 6$ , we note that the RHS is an integer, the LHS must also be an integer. Since $\lfloor x^2 \rfloor$ is an integer, $x \lfloor x \rfloor$ must also be an integer. For $x \lfloor x \rfloor$ to be an integer, $x$ must be of the form $x = \lfloor x \rfloor + \{x\} = \lfloor x \rfloor + \frac n{\lfloor x \rfloor}$ , where $n$ is an integer such that $0 \le n < \lfloor x \rfloor$ .

Now we have:

$\begin{aligned} \left \lfloor x^2 \right \rfloor - x \lfloor x \rfloor & = 6 \\ \left \lfloor \left(\lfloor x \rfloor + \{x\} \right)^2 \right \rfloor - \left(\lfloor x \rfloor + \{x\} \right) \lfloor x \rfloor & = 6 \\ \left \lfloor \left(\lfloor x \rfloor + \frac n{\lfloor x \rfloor} \right)^2 \right \rfloor - \left(\lfloor x \rfloor + \frac n{\lfloor x \rfloor} \right) \lfloor x \rfloor & = 6 \\ \left \lfloor \lfloor x \rfloor^2 + 2n +\frac {n^2}{\lfloor x \rfloor^2} \right \rfloor - \lfloor x \rfloor^2 - n & = 6 \\ \lfloor x \rfloor^2 + 2n - \lfloor x \rfloor^2 - n & = 6 \\ \implies n & = 6 \end{aligned}$

Therefore, the equation holds true when $n=6$ , Since $n < \lfloor x \rfloor$ , there is no solution for $\lfloor x \rfloor \le 6$ ; and the smallest $\lfloor x \rfloor = 7$ and the smallest $x = 7 + \dfrac 67 = \dfrac {55}7$ . $\implies a + b = 55 + 7 = \boxed{62}$

Let x = a + b, where a is a natural number and 0<=b<1. So, [x] (I'm using brackets to represent the floor function because I don't know how to use latex. My apologies.) equals a.

X[x] = (a)(a + b) = a^2 + ab, where a^2 is necessarily an natural number, and nothing is known about ab (it can be a fraction between 0 and 1, a natural number, or a fraction greater than 1)

x^2, then, equals a^2 + 2ab + b^2, where b^2 is necessarily between 0 and 1, a^2 is necessarily a natural number, and nothing is known about 2ab (it can be a fraction between 0 and 1, a natural number, or a fraction greater than 1). So, [x^2] = a^2, or a^2 + 2ab, or something in between a^2 and a^2 + 2ab.

This means that [x^2] - x[x] equals -ab, +ab, or something in between -ab and +ab. Since this expression equals 6 and x is positive, the expression must be somewhere between 0 and ab. Since we want a to be as small as possible we want [x^2] - x[x] to be as big as possible, so we now know that [x^2] - x[x] must equal ab.

Now we have ab = 6. So b = 6/a. And since b is between 0 and 1, the smallest value of a for where 6/a < 1 is a = 7.

We're almost done! a = 7 and b = 6/7, so x = a + b = 55/7. And 55 + 7 = 62.

Write $x = a + \epsilon$ with $0 \leq \epsilon < 1$ ; and $x^2 = a^2 + b + \eta$ with $b \geq 0$ an integer and $0 \leq \eta < 1$ . The equation becomes $a^2 + b - (a+\epsilon)a = 6\ \ \ \ \therefore\ \ \ b = 6 + a\epsilon.$ Substitute this in the equation for $x^2$ : $6 + \eta = a\epsilon + \epsilon^2.$ This means that $a > 6$ . Because we look for the minimum value of $x$ , we choose $a = 7$ .

Now $b = 6 + 7\epsilon$ is an integer, so that $\epsilon = k/7$ , with $k$ an integer. We get $6 + \eta = k + \frac{k^2}{49} \ \ \ \therefore\ \ \ k^2 + 49k = 294 + 49\eta.$ Now $0 \leq \eta < 1$ , so that $294 \leq k^2 + 49k < 343, \\ 294 \leq (k + 24\tfrac12)^2 - 306\tfrac14 < 343, \\ \sqrt{894\tfrac14} \leq \left|k + 24\tfrac12\right| < \sqrt{943\tfrac14}, \\ 29.90\cdots \leq \left|k + 24\tfrac12\right| < 30.71\dots.$ The only positive integer solution is $k + 24\tfrac12 = 30\tfrac12\ \ \ \therefore\ \ \ k = 6.$ We now have the solution $\epsilon = 6/7$ and $x = 7+\frac 6 7 = \boxed{\dfrac{55}{7}}.$

$\big\lfloor \left(\frac{55}7\right)^2\big\rfloor-\frac{55}7\big\lfloor\frac{55}7\big\rfloor=6$

We can see that this is true by doing the individual operations:

$\big\lfloor \left(\frac{55}7\right)^2\big\rfloor=61$

$\frac{55}7\big\lfloor\frac{55}7\big\rfloor=\frac{55}7\times7=55$

$61-55=6$

Answer $=\boxed{\dfrac{55}7}$