Brazilian Math Olympiad

Consider the number $\underset{n\text{ times}}{\underbrace{55\dots 5}}$. The sum of the squares of its digits is $n\cdot 5^2=25n$. We can exchange any two fives by one three and one four, so the sum of the squares decreases by $5^2$, until we run out of fives. So we can get any sum from $25\cdot \lfloor n/2\rfloor$ and $25\cdot n$. So it suffices to show that there is an integer $k$ such that $\frac{n}{2}\le k^2\le n$. Choose $k$ such that $k^2\le n<(k+1{)}^2$. Suppose $k^2<\frac{n}{2}$. Then $n>2k^2$, and $(k+1{)}^2>n>2k^2⟹(k+1{)}^2\ge 2k^2+2⟺k^2-2k+1\le 0⟺(k-1{)}^2\le 0$, which is false except for $k=1$, or $2<n<4$, that is, $n=3$. But the statement of the problem itself gives an example with $n$ digits: $122$.