Sums of Partial Sequences of Squares

Theorem
Let $n \in \Z_{>0}$.

Consider the odd number $2 n + 1$ and its square $\paren {2 n + 1}^2 = 2 m + 1$.

Then:
 * $\ds \sum_{j \mathop = 0}^n \paren {m - j}^2 = \sum_{j \mathop = 1}^n \paren {m + j}^2$

That is:
 * the sum of the squares of the $n + 1$ integers up to $m$

equals:
 * the sum of the squares of the $n$ integers from $m + 1$ upwards.

Proof
First it is worth rewriting this so as to eliminate $m$.

Thus the statement to be proved can be expressed:


 * $\ds \sum_{j \mathop = 0}^n \paren {2 n^2 + 2 n - j}^2 = \sum_{j \mathop = 1}^n \paren {2 n^2 + 2 n + j}^2$

We have:

Hence the result.