Czech-Polish-Slovak Mathematical Match

We prove that $c$ satisfies the condition in the statement if and only if $c\le 2$.

Let us first consider any $c\le 2$. For any positive integer $k$, define $$n=k^2+1\text{and}m=(k-1{)}^2+1$$ Observe that $$m+c\sqrt{m-1}+1\le k^2-2k+2+2(k-1)+1=k^2+1=n$$ and $$(n,n+1,\dots ,2n-m)=(k^2+1,k^2+2,\dots ,k^2+2k).$$ Therefore, every such pair $(n,m)$ indeed satisfies the property from the problem statement, and there are infinitely many such pairs.

Now let us consider any $c>2$, and let $(n,m)$ be any pair of positive integers satisfying the property from the problem statement. Observe that for each positive integer $n$, the number $\lceil \sqrt{n}{\rceil }^2$ is always between numbers $n$ and $(\sqrt{n}+1{)}^2$ (inclusive), hence there is always a square of an integer in the range $$n,n+1,\dots ,n+\lfloor 2\sqrt{n}\rfloor +1.$$ This implies that $2n-m<n+\lfloor 2\sqrt{n}\rfloor +1$, so in particular $$m\ge n-2\sqrt{n}.$$ Combining this with the inequality from the problem statement yields $$n\ge n-2\sqrt{n}+c\sqrt{n-2\sqrt{n}-1}+1.(1)$$ Observe that since $c>2$, we have $c\sqrt{n-2\sqrt{n}-1}>2\sqrt{n}$ for large enough $n$. Indeed, equivalently we have $1-\frac{2}{\sqrt{n}}-\frac{1}{n}>\frac{4}{c^2}$, and the left-hand side tends to 1 as $n$ grows to infinity while the right hand side is strictly smaller than 1. This implies that (1) may be satisfied only for finitely many positive integers $n$. Since $m\le n$ for all pairs $(n,m)$ satisfying the conditions from the problem statement, this implies that there are only finitely many such pairs $(n,m)$. $\square$