Belarusian Mathematical Olympiad

a) The solution of part a) is almost obvious and its statement holds for any $b\in \mathbb{Q}$ and $a\in \mathbb{N}$. Indeed, if $n\sqrt{n^2+a}=bm$, then, since $b\in \mathbb{Q}$, the number $\sqrt{n^2+a}$ must be rational and therefore integer. Hence $n^2+a=k^2$ for some positive integer $k$. But the differences between consecutive perfect squares infinitely increase, there is only finite number of pairs $(n,k)$ of positive integers $n$ and $k$ such that for the given number $a$ holds $n^2-k^2=a$.

b) Let $b=p/q$ be the irreducible fraction representation of $b$, according to conditions of the problem $a$ and $q$ are odd. Denote by $I_n$ the set of all positive integers from interval $[q{n}^2,q{n}^2+aq/2]$. Since $a$ and $q$ are odd, $I_n=\{q{n}^2+i:i=0,(aq-1)/2\}$. Consider another set $P=\{(n,m):n\in \mathbb{N},pm\in I_n\}$. Substituting $b=p/q$ to the expression of $M(n,m)$, we get $M(n,m)=q^{-1}|qn\sqrt{n^2+a}-pm|$. From the obvious chain of inequalities $$n^2<n\sqrt{n^2+a}<n^2+a/2(\ast )$$ it follows that $q{n}^2<qn\sqrt{n^2+a}<q{n}^2+(aq)/2$. Therefore, if positive integer $mp$ doesn't belong to $I_n$, then $M(n,m)\ge q^{-1}(1-\{aq/2\})=1/(2q)$ (where $\{\cdot \}$ denotes the fractional part, and since $aq$ is odd, $\{aq/2\}=1/2$). Hence $$\inf \{M(n,m):(n,m)\notin P\}\ge 1/(2q),(\ast \ast )$$ and the zeroes of the function $M(n,m)$ are contained in $P$. Let us find the lower bound of $M(n,m)$ for $(n,m)\in P$ for all $n$ large enough. If $(n,m)\in P$, then $pm=q{n}^2+i$, where $i\in \{0,1,\dots ,(aq-1)/2\}$. For such $n$ and $m$ we get $$M(n,m)=q^{-1}\left|qn\sqrt{n^2+a}-q{n}^2-i\right|=\frac{(q^{2}a-2qi)n^2-i^2}{q(qn\sqrt{n^2+a}+q{n}^2+i)}.$$ The resulting fraction has the least numerator and the largest denominator if $i$ is maximal, i. e. $i=(aq-1)/2$. Substituting this value of $i$ and taking into account the right inequality of $(\ast )$, we obtain that $$M(n,m)\ge \frac{n^2-(aq-1{)}^2/4}{2q^2{n}^2+a{q}^2-q/2}\to \frac{1}{2q^2}\text{as}n\to \infty ,$$ so, there exists an $n_0$ such that $M(n,m)>1/(4q^2)$ for all $(n,m)\in P$ with $n\ge n_0$. In particular, because if $(\ast \ast )$ holds then this means that the zeroes of $M(n,m)$ belong to the finite set $P_0=\{(n,m)\in P:n\ge n_0\}$. Therefore $M((n,m))$ has the finite number of zeroes. Choose the minimal nonzero value of $M(n,m)$ on the set $P_0$ (say $C_0$). Then we can take $C=\min \{C_0,1/(4q^2)\}$ as the required constant. The statement of part b) is proved.