Q:

IMO 1992 Problem 6

For each positive integer $\,n,\;S(n)\,$ is defined to be the greatest integer such that, for every positive integer $\,k\leq S(n),\;n^{2}\,$ can be written as the sum of $\,k\,$ positive squares. 

a.) Prove that $\,S(n)\leq n^{2}-14\,$ for each $\,n\geq 4$.
Now solution says for part a)
Representing $n^2$ as a sum of $n^2-13$ squares is equivalent to represent $13$ as sum of numbers of the form $x^2-1$,..
I did not get why it is equivalent ??? 

A:

$$\begin{align}
n^2&=\sum_{k=1}^{n^2-13}a_k^2\\
&=\sum_{k=1}^{n^2-13}(a_k^2-1)+(n^2-13)
\end{align}$$
Therefore,
$$13=\sum_{k=1}^{n^2-13}(a_k^2-1)$$

