Czech-Slovak-Polish Match

The numbers $a=(k-1)k$, $b=(k+1)k$, $c=(k-1)(k+1)$, $d=k^2$ clearly satisfy the equality $ab=cd$ and the inequalities $a<c<d<b$ for any $k>1$. Let thus $k$ be the least natural number for which $p<a$, i.e. $p<(k-1)k$ (for a given $p$). We will show that for this $k$ necessarily $b=(k+1)k\le p+4+2\sqrt{4p+1}$, which is evidently a number by $\frac{1}{4}$ smaller than the upper bound of the interval in our problem, so we will be done.

In view of the choice of the number $k$ we have $p\ge (k-2)(k-1)$. Solving this quadratic inequality yields the estimate $$k\le \frac{3}{2}+\sqrt{p+\frac{1}{4}},$$ from which it already follows that $$\begin{gathered} b=(k+1)k\le \left(\frac{5}{2}+\sqrt{p+\frac{1}{4}}\right)\cdot \left(\frac{3}{2}+\sqrt{p+\frac{1}{4}}\right) \\ =\frac{15}{4}+4\sqrt{p+\frac{1}{4}}+\left(p+\frac{1}{4}\right)=p+4+2\sqrt{4p+1}. \end{gathered}$$