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
 A: 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 $n=k^2$, $$x=\frac94\quad\text{and}\quad x=k^2+\frac2k+\frac1{k^4},\quad k\in\mathbb N,\ k>2.$$
A: There is no solution in $(3,4)$ because it has to check
$$3(\sqrt{x}-1)=1,$$
which is true for $x=16/9$.
You need to check the solutions in the intervals $(1,4)$, $(4,9)$, $(9,16)$,... and isolate the $x$ in the produced equations : in $(n^2,(n+1)^2)$, the equations are
$$m(\sqrt{x}-n)=1,$$
with $n^2\leq m \leq (n+1)^2-1.$
A: 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.
