Solve for x: $\lfloor x\rfloor \{\sqrt{x}\}=1$ where $\{x\}=x-\lfloor x\rfloor$ Solve for x:
$\lfloor x\rfloor \{\sqrt{x}\}=1$
where $\{x\}=x-\lfloor x\rfloor$
My Attempt:
I took intervals of $x$ as $x\in (2,3)$ so $\sqrt x\in (1,2)$. Due to which $\lfloor x\rfloor=2$ and $\{\sqrt x\}=\sqrt x-\lfloor \sqrt x\rfloor=\sqrt x-1$.
Substituting in the equation
$\lfloor x\rfloor \{\sqrt{x}\}=1$
I got
$2(\sqrt x-1)=1$
$x=\frac{9}{4}$
which satisfies the equation.
But when I apply the same process to $x\in (3,4)$ I end up getting value of $x$ that does not satisfy the given equation. But when I plotted the graph a solution clearly exists between two consecutive integers.
What am I missing

There is obviously no solution for x < 1. Let k, n be integers with k ≥ 1 and $k^2 ≤ n ≤ k^2 + 2k$, and look for solutions x in the interval [n, n+1); this covers all solutions.
In this interval, the equation is equivalent to $n \cdot (\sqrt x - k) = 1$ or $x = (k + 1 / n)^2$. Since $n ≥ k^2$, $x ≤ (k + 1 / k^2)^2 = k^2 + 2/k + 1/k^4$. For k ≥ 3, the right hand side is less than $k^2 + 1$, so we have $n = k^2$ and $x = (k + 1 / k^2)^2$.
For k = 2, n ≥ 5 is also impossible because in that case $x = (k + 1 / n)^2 ≤ (2 + 1/5)^2 = 121/25 < 5$. For k = 1, $x = (1 + 1 / n)^2$, so if n = 1 then x = 4, if n = 2 then x = 2.25, if n = 3 then x = 16/9, and x = 2.25 is the only solution.
Summary: The solutions are x = 2.25, and $x = (k + 1 / k^2)^2$ for k ≥ 2. To check, let k = 10, $x = 10.01^2$, $\sqrt x = 10.01$, and 100 * 0.01 = 1.
