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
The proof proceeds by induction.

For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
 * $\ds \sum_{j \mathop = 0}^n \paren {m - j}^2 = \sum_{j \mathop = 1}^n \paren {m + j}^2$

where $\paren {2 n + 1})^2 = 2 m + 1$.

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

Thus the statement to be proved can be expressed:


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

Basis for the Induction
$\map P 1$ is the case:
 * $3^2 + 4^2 = 5^2$

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

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

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

So this is the induction hypothesis:
 * $\ds \sum_{j \mathop = 0}^k \paren {2 k^2 + 2 k - j}^2 = \sum_{j \mathop = 1}^k \paren {2 k^2 + 2 k + j}^2$

from which it is to be shown that:
 * $\ds \sum_{j \mathop = 0}^{k + 1} \paren {2 \paren {k + 1}^2 + 2 \paren {k + 1} - j}^2 = \sum_{j \mathop = 1}^{k + 1} \paren {2 \paren {k + 1}^2 + 2 \paren {k + 1} + j}^2$

Induction Step
This is the induction step:

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

Therefore:
 * $\ds \forall n \in \Z_{>0}: \sum_{j \mathop = 0}^n \paren {m - j}^2 = \sum_{j \mathop = 1}^n \paren {m + j}^2$

where $\paren {2 n + 1}^2 = 2 m + 1$.