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

We can proceed similar as you did. Let $n=\lfloor x\rfloor$. From the assumption $\lfloor x\rfloor\{\sqrt x\}=1$ one concludes $\{\sqrt x\}=\frac1n$. Since $\sqrt x$ changes its integral part at integral values, we can conclude that $\lfloor\sqrt n\rfloor=\lfloor\sqrt x\rfloor$, so that
$$\sqrt x=\lfloor\sqrt x\rfloor+\{\sqrt x\}=\lfloor\sqrt n\rfloor+\frac1n\Longrightarrow x=\lfloor\sqrt n\rfloor^2+\frac{2\lfloor\sqrt n\rfloor}{n}+\frac1{n^2}.$$
Now we have to check under which conditions this value for $x$ indeed satisfies the assumption $\lfloor x\rfloor=n$. For this, let $n=k^2+m$ with $k,m\in\mathbb N_0$ and $k$ maximal, i.e. $\lfloor\sqrt n\rfloor=k$. Now if $n\geq5$, then
$$x=k^2+\frac{2k}{n}+\frac1{n^2}=k^2+\frac{2kn+1}{n^2}<k^2+1,$$
so in this case, we must have $m=0$. Then we have $n=k^2<x<k^2+1$, so indeed $\lfloor x\rfloor=n$, hence we found solutions. The cases $n=0,\ldots,4$ are easily checked, and $\frac94$, which you found, is the only solution in these cases.
To summarize, the solutions are $\frac94$ and $x=\lfloor\sqrt n\rfloor^2+\frac{2\lfloor\sqrt n\rfloor}{n}+\frac1{n^2}$ with $n$ a perfect square other than $1$ or $4$, or letting $n=k^2$, $$x=\frac94\quad\text{and}\quad x=k^2+\frac2k+\frac1{k^4},\quad k\in\mathbb N,\ k>2.$$
