Sums of Partial Sequences of Squares

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

Consider the odd number $2 n + 1$ and its square $\left({2 n + 1}\right)^2 = 2 m + 1$.

Then:
 * $\displaystyle \sum_{j \mathop = 0}^n \left({m - j}\right)^2 = \sum_{j \mathop = 1}^n \left({m + j}\right)^2$

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
The proof proceeds by induction.

For all $n \in \Z_{> 0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \sum_{j \mathop = 0}^n \left({m - j}\right)^2 = \sum_{j \mathop = 1}^n \left({m + j}\right)^2$

where $\left({2 n + 1}\right)^2 = 2 m + 1$.

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

Thus the statement to be proved can be expressed:


 * $\displaystyle \sum_{j \mathop = 0}^n \left({2 n^2 + 2 n - j}\right)^2 = \sum_{j \mathop = 1}^n \left({2 n^2 + 2 n + j}\right)^2$

Basis for the Induction
$P \left({1}\right)$ is the case:
 * $3^2 + 4^2 = 5^2$

which is the $3-4-5$ triangle.

Thus $P \left({1}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is the induction hypothesis:
 * $\displaystyle \sum_{j \mathop = 0}^k \left({2 k^2 + 2 k - j}\right)^2 = \sum_{j \mathop = 1}^k \left({2 k^2 + 2 k + j}\right)^2$

from which it is to be shown that:
 * $\displaystyle \sum_{j \mathop = 0}^{k + 1} \left({2 \left({k + 1}\right)^2 + 2 \left({k + 1}\right) - j}\right)^2 = \sum_{j \mathop = 1}^{k + 1} \left({2 \left({k + 1}\right)^2 + 2 \left({k + 1}\right) + j}\right)^2$

Induction Step
This is the induction step:

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall n \in \Z_{>0}: \sum_{j \mathop = 0}^n \left({m - j}\right)^2 = \sum_{j \mathop = 1}^n \left({m + j}\right)^2$

where $\left({2 n + 1}\right)^2 = 2 m + 1$.