# Rational solutions

Find $$x \in \mathbb{Q}$$ such that $$1 + \frac{2}{[x]} + \frac{3}{{x - [x] }} = 4 {x}$$. Initially I was thinking that is easy. If we separate the therms with integer part and fractionary part and consider the functions with fractionary part she is continous functions and we have a solution all the time but we don't know if it is rational. I was trying to consider and an equation of degree two but doesn't work.

If we write $$k = [x]$$, then the equation translates to $$1 + \frac 2 k + \frac 3 {x - k} = 4x$$, which after simplification gives $$4k x^2 - (4k^2 + k + 2)x + k^2 - k = 0.$$ Since this quadratic equation in $$x$$ has a rational solution, its discriminant $$\Delta$$ must be the square of some rational number $$r$$ (note that $$k \neq 0$$).

Thus we write $$r^2 = \Delta = (4k^2 + k + 2)^2 - 4 \cdot 4k \cdot (k^2 - k) = 16k^4 - 8k^3 + 33k^2 + 4k + 4.$$ Since $$\Delta$$ is an integer, we know that $$r$$ is also an integer, which we shall assumed to be non-negative.

Completing the square, we get $$r^2 = (4k^2 - k + 4)^2 + 12k - 12$$.

There are three cases: $$k = 1$$, $$k > 1$$, $$k < 1$$.

For $$k = 1$$, we solve the equation and get $$x = 0$$ or $$\frac 7 4$$. Obviously $$x = \frac 7 4$$ is a possible solution to the original equation.

If $$k > 1$$, then we have $$r^2 > (4k^2 - k + 4)^2$$ which gives $$r \geq 4k^2 - k + 5$$ (note that $$4k^2 - k + 4$$ is strictly positive for any $$k$$). Thus we have $$(4k^2 - k + 5)^2 \leq r^2 = (4k^2 - k + 4)^2 + 12k - 12$$ which leads to $$8k^2 - 14k + 21 \leq 0$$, which is impossible.

If $$k < 1$$, then a similar argument gives $$r \leq 4k^2 - k + 3$$ and hence $$(4k^2 - k + 3)^2 \geq r^2 = (4k^2 - k + 4)^2 + 12k - 12$$ which leads to $$8k^2 + 10k - 5 \leq 0$$. It is easy to see that the only integers $$k$$ satisfying this inequality are $$k = 0, -1$$, and solving the original equation confirms that no rational solution $$x$$ exists in this case.

Therefore the only rational solution is $$x = \frac 7 4$$.