National problem · Algebra

For each \(n = 1, 2, 3, \ldots\) Perica looks for the smallest block of \(2n+1\) consecutive positive integers with the property that the sum of the squares of the smallest \(n+1\) of them equals the sum of the squares of the remaining \(n\). If such a block exists, he writes it into row \(n\) of his pyramid; if it does not, row \(n\) stays empty. After the first three steps \((n = 1\), \(n = 2\), \(n = 3)\) his pyramid looks like this:

\begin{aligned} 3^2 + 4^2 &= 5^2 \\ 10^2 + 11^2 + 12^2 &= 13^2 + 14^2 \\ 21^2 + 22^2 + 23^2 + 24^2 &= 25^2 + 26^2 + 27^2 . \end{aligned}

We say that the numbers \(3, 4, 5\) lie in the first row; that \(10, 11, 12, 13, 14\) lie in the second row; and that \(21, 22, 23, 24, 25, 26, 27\) lie in the third row. If Perica goes on building his pyramid in this way, will the number \(2011\) ever appear in one of the rows?