<s>[INST] Sum of odd numbers always gives a perfect square. $1 + 3 = 4$ (or $2$ squared)
$1+3+5 = 9$ (or $3$ squared)
$1+3+5+7 = 16$ (or $4$ squared)
$1+3+5+7+9 = 25$ (or $5$ squared)
$1+3+5+7+9+11 = 36$ (or $6$ squared)
you can go on like this as far as you want, and as long as you continue to add odd numbers in order like that, your answer is always going to be a perfect square.
But how to prove it?
 [/INST] Because
$$\begin{align}1+3+5+\cdots+(2n-1)&=\sum_{k=1}^{n}(2k-1)\\&=2\sum_{k=1}^{n}k-\sum_{k=1}^{n}1\\&=2\cdot\frac{n(n+1)}{2}-n\\&=n(n+1)-n\\&=n^2.\end{align}$$
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