IMO 1992 Problem 6

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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 ???

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