Find the positive solutions for $x + 2 \{ x \} = 3 \lfloor x \rfloor$

Question

Find the positive solutions for $x + 2 \{ x \} = 3 \lfloor x \rfloor$

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attempt:

Notice that the equation can be rewritten as

$$ x + 2 \{ x\} = 2 \lfloor x \rfloor + x - \{x\}$$ $$ 3 \{x\} = 2 \lfloor x \rfloor $$

From this we know that $3 \{x\}$ must be even positive integer. But the only $\{x\}$ that makes it integer is $\{ x\} = 1/3$ (but $3\{x\}$ is odd) and $\{x\} = 0$. So there is no positive integer solution?

Answer 1

Hint:

You forgot about $\{x\}=2/3$.

Answer 2

Right. So $\{x\} = \frac k{3}$ where $k= 2 \lfloor x \rfloor$, an integer.

As $0\le \{x\}=\frac k3 < 1$ we have $k=2\lfloor x \rfloor = 0, 1,2$ but $k$ is even, so $k = 2j = 0, 2; j=\lfloor x \rfloor = 0,1$.

And $x = \lfloor x \rfloor +\{x\} = j + \frac {2j}3$ where $j = 0,1$.

And that's it. All solution, negative, positive, or zero.

Answer 3

A more transparent way:

We can write $x$ as $x=n+r$, where $n=\lfloor x \rfloor; ~n \in 0,1,2.. $ and $r=\{x\};~ 0 \le r < 1. $

Equation reads as,

$n+r+2r=3n \Rightarrow r= \frac{2n}{3} \Rightarrow r =0 , \frac {2}{3}.$

so $x=0, \frac{5}{3}$