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 2 of 4: At most three solutions

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r-3<k\le r\ \Rightarrow\ \text{at most 3 values of } k

Detailed analysis

Dividing $k^2\le rk$ by $k$ gives $k\le r$ . Since $k\ge1$ , $2k+1\le3k$ , so $rk<(k+1)^2=k^2+2k+1\le k^2+3k$ gives $r<k+3$ . Hence every valid $k$ lies in the half-open interval $(r-3,r]$ , which contains exactly three integers, so there are at most three possible values of $k$ ; since $x=\sqrt{rk}$ is strictly increasing in $k$ , different valid $k$ give different $x$ , so there are at most three positive solutions $x$ .