Problem 1

Prove that for each real number

r>2

, there are exactly two or three positive real numbers

x

satisfying the equation

x^2=r\lfloor x\rfloor

. (Here

\lfloor x\rfloor

denotes the largest integer less than or equal to

x

Step 3 of 4: At least two solutions

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k\in\{\lfloor r\rfloor,\lfloor r\rfloor-1\}\ \Rightarrow\ x=\sqrt{rk}\ \text{is a genuine solution}

Detailed analysis

Conversely take any integer $k$ with $r-2<k\le r$ ; there are always exactly two such integers, namely $k=\lfloor r\rfloor$ and $k=\lfloor r\rfloor-1$ (both positive since $r>2$ ). For such $k$ , $k\le r$ gives $rk\ge k^2$ , and $k>r-2$ gives $r<k+2$ , hence $rk<(k+2)k<(k+1)^2$ (as $k\ge1$ gives $(k+2)k=k^2+2k<k^2+2k+1$ ). So $k^2\le rk<(k+1)^2$ , i.e. $k\le\sqrt{rk}<k+1$ , so $x=\sqrt{rk}$ satisfies $\lfloor x\rfloor=k$ and is a genuine solution of $x^2=r\lfloor x\rfloor$ .