Sum of odd numbers always gives a perfect square.

Consider a square. It can be divided as the following:

Each red segment contains odd number of circles.

\begin{align*} \sum_{k=0}^{n}[2k+1] &= \sum_{k=0}^{n}[2k] + \sum_{k=0}^{n}1=\sum_{k=0}^{n}[2k] + (n+1)\\ &=2\sum_{k=0}^{n}[k] + (n+1)= 2\frac{n(n+1)}{2}+(n+1) = (n+1)^2 \end{align*}

Because $$\begin{align}1+3+5+\cdots+(2n-1)&=\sum_{k=1}^{n}(2k-1)\\&=2\sum_{k=1}^{n}k-\sum_{k=1}^{n}1\\&=2\cdot\frac{n(n+1)}{2}-n\\&=n(n+1)-n\\&=n^2.\end{align}$$