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Solve for $x\in \mathbb{R} $ the equation $$ \left\{ \frac{\left[ x\right] +1}{2014}\right\} =2014x+\frac{1}{2014}, $$ where $\left\{ x\right\}$ is the fractional part of x and the $[x]$ is the integer part of x.$$$$ I took $x=0$ and the equation is verified, but I don't know how to show that this the only one...

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We know that $\left\{x\right\}\in[0,1),\forall x\in\mathbb{R}$, hence $$ 0\le 2014x+\frac{1}{2014}<1 \implies0\le 2014^2x+1<2014. $$ From the last inequality we obtein that $x\ge 0$ and $$ x<\frac{2013}{2014^2}<\frac{2014}{2014^2}<\frac{1}{2014}\implies [x]=0.$$The equation begins $$\left\{\frac{1}{2014}\right\}=2014x+\frac{1}{2014}\implies2014x=0\implies x=0.$$

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