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 1 of 4: Set up via the floor value

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k=\lfloor x\rfloor,\quad x^2=rk,\quad k\le x<k+1

Detailed analysis

Let $x>0$ satisfy $x^2=r\lfloor x\rfloor$ and set $k=\lfloor x\rfloor$ . Since $x>0$ and $r>0$ , $x^2=rk>0$ forces $k$ to be a positive integer, and $k\le x<k+1$ gives $k^2\le x^2=rk<(k+1)^2$ .