You Can't Get An Equation Shorter Than This.

Write $x = n+r$ with $n \in \mathbb Z$ and $0 \le r < 1$ . Then $rn = 2014(r+n),$ so $(2014-r)(2014-n) = 2014^2$ , so $2014 - n$ is between $2014$ and $2014^2/2013$ , so $n$ is between $-2014/2013$ and $0$ . So $n = -1$ or $n = 0$ .

If $n = 0$ then $x = 0$ . If $n = -1$ then $r = 2014/2015$ , so $x = -1/2015$ . So the sum of these two solutions is $-1/2015$ , so the answer is $\fbox{2016}$ .

Firstly, $x=0$ is clearly a solution to the equation.

Now let $\lfloor x \rfloor = y$ and $\left\{x\right\} = z$

So $x = y+z$

$yz = 2014x = 2014\left(y+z\right)$

$\Rightarrow yz = 2014y + 2014z$

$yz - 2014z = 2014y$

$z = \frac{2014y}{y-2014}$

$0 \leq z < 1 \Rightarrow 0 \leq \frac{2014y}{y-2014} < 1$

Note also that if we assume $x>0$ , then $y < x$ and $z<1$ so $yz < x \Rightarrow yz \neq 2014x$

This means that if there is indeed a non-zero solution to the equation, then $x$ cannot be positive, and thus $x,y <0$

Now we see that $y-2014 < 0 \Rightarrow y-2014 < 2014y \leq 0$

Taking the left half we see that $-2013y < 2014 \Rightarrow y> \frac{-2014}{2013}$

$\Rightarrow 0 > y > \frac{-2014}{2013}$

$y \in \mathbb{Z} \Rightarrow y = -1$

$yz = 2014y + 2014z \Rightarrow y = \frac{2014z}{z-2014}$

$2014-z = 2014z \Rightarrow z = \frac{2014}{2015}$

So the sum of solutions is then $0 + \left(y+z\right) = 0 + -1 + \frac{2014}{2015} = \frac{1}{2015}$

$1 + 2015 = 2016$

Let $x:=m + \varepsilon$ where $m \in \mathbb{Z}$ and $\varepsilon \in \langle 0,1)$ . Thus $m\varepsilon = 2014(m+\varepsilon)$ . Let express $\varepsilon = \varepsilon(m) = \frac{2014m}{m-2014}$ , epsilon as function of m. To satisfy assumption $\varepsilon \in \langle 0,1)$ :

$\varepsilon(0) = 0$ thus $x_1 = 0$ .

$\varepsilon(1) = -\frac{2014}{2013} < 0$ , for any $|m| > 1$ : $\varepsilon$ is either negative or bigger than $1$ (numerator increases linearly with coeff $2014$ , denominator linearly with coeff $1$ ).

$\varepsilon(-1) = \frac{2014}{2015}$ thus $x_2 = -1 + \frac{2014}{2015} = -\frac{1}{2015}$ and $\boxed{ a+b = 2016}.$

Begin from the end:

(P, Q) = (0, 0) and (- 0, - 4.962779156327543424317617866005e-4)

x1 = 0

x2 = - 4.962779156327543424317617866005e-4

x1 + x2 = - 4.962779156327543424317617866005e-4 = - 1/ 2015 = - a/ b

a + b = 2016

Check with x = -1/ 2015,

(x - Floor(x)) Floor(x) = 2014 x

(-1/ 2015 + 1)(-1) = 2014 (-1/ 2015)

-0.9995037220843672456575682382134 =

-0.9995037220843672456575682382134

Working:

Int p * Frac p = 2014 (Int p + Frac p) ELSE (Int -p - 1)(Frac -p + 1) = 2014 (Int -p + Frac -p)

P Q = 2014 (P + Q) OR (- P - 1)(- Q + 1) = - 2014 (P + Q)

Q = 2014 + 2014^2/ (P - 2014) OR Q = - 2013 + (2013*2015 + 1)/ (P + 2015)

Search with positive integers P from 0 to extreme only, for Q > 0 and ABS(Q) < 1 as (P + Q) or -(P + Q) are combination of whole number and fractional decimals only.

Only (0, 0) and (0, - 1/ 2015) are solutions.