55rd Ukrainian National Mathematical Olympiad - Third Round (Second Tour)

Suppose the opposite. Let $$m^2+a\in [m^2-2m,m^2+2m]\text{ i }{m}^4+1\equiv m^2+a.$$ Since $$\begin{gathered} (m^4+1,m^2+a)=(-a{m}^2+1,m^2+a)=(a^2+1,m^2+a),\text{ then }{a}^2+1\equiv m^2+a, \\ {a}^2+1=(m^2+a)r. \end{gathered}$$ Let $a\in [-2m+3,0]$, then $\frac{m^4+1}{m^2+a}\le \frac{m^4+1}{m^2-2m+3}<m^2+2m+1$, so $m^4+1$ has a divisor that is no less than $m^2$ and fulfills the assumption, therefore instead of $m^2+a$ we can examine a divisor $m^2+b$, where $b\ge 0$. If $a=-2m+2$, then $4m^2-8m+5\equiv m^2-2m+2$, so $3\equiv m^2-2m+2$, which is impossible in case of $m>1$. If $a=-2m+1$, then $4m^2-4m+2\equiv (m-1{)}^2-m-1$, thus $2\equiv m-1$, $m\in \{2,3\}$. In case of $m=2$ and $m=3$: $m^4+1=17$ and $m^4+1=2\cdot 41$ - none of these numbers fulfills the condition. If $a=-2m$, then $1\equiv m$ is a contradiction. Thus, if number $m^4+1$ has a divisor $m^2+a$, where $a\in [-2m,2m]$, then the number $m^4+1$ has a divisor $m^2+b$, where $b\in [0,2m]$. It is also clear that $b\ne 0$. In such case, $4m^2+1\ge b^2+1=(m^2+b)r\ge m^{2}r$, so $r\le 3$. Case $r=3$: $b^2+1=3(m^2+b)$ - it is impossible, because $b^2+1$ is not divisible by 3. Case $r=2$: $b^2+1=2(m^2+b)$, $(b-1{)}^2=2m^2$ - impossible. Case $r=1$: $b^2-b+1=m^2$, but $b^2>b^2-b+1=m^2>(b-1{)}^2$, so this case is also impossible. Thus, we have the contradiction with the supposition, so the statement is proved.