Odd Square Sum

Note that a perfect square is always of the form $4k$ or $4k+1$ . (Consider squares of odd and even numbers)

Now, sum of square of two odd numbers is,

$(2m+1)^2 + (2n+1)^2 = (4m^2+4m+1)+(4n^2+4n+1) = 4k+2$ , which cannot be a perfect square.

Likewise is not true for the other 3 cases.

$(n+1)^2=n^2+(2n+1)$

meaning that $(n+1)^2$ (a square), can be written as the sum of another square $n^2$ plus $2n+1$ of $1^2$ (another square). So, if we take $n$ to be an odd number, we have found a group of $2n+2=2(n+1)$ odd squares that sum up to a square. For $n=1,3,5,7$ , the length of the sum would be $4,8,12,16$ respectively and it would cover all the options, except $n=2$ , which is actually well-known due to Pythagorean triples.