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. 

[b]a.)[/b] Prove that $\,S(n)\leq n^{2}-14\,$ for each $\,n\geq 4$. 
[b]b.)[/b] Find an integer $\,n\,$ such that $\,S(n)=n^{2}-14$. 
[b]c.)[/b] Prove that there are infintely many integers $\,n\,$ such that $S(n)=n^{2}-14.$
### Part (a)
1. We need to prove that \( S(n) \leq n^2 - 14 \) for each \( n \geq 4 \).
2. Assume for contradiction that \( S(n) > n^2 - 14 \). This means there exists \( k = n^2 - 13 \) such that \( n^2 \) can be written as the sum of \( k \) positive squares.
3. Consider the set with \( n^2 - 13 \) elements \(\{1^2, 1^2, \ldots, 1^2\}\). The sum of these elements is \( n^2 - 13 \).
4. To reach \( n^2 \), we need to increase the sum by 13. We can do this by changing some of the \( 1^2 \) terms to larger squares.
5. Changing one \( 1^2 \) to \( 4^2 \) increases the sum by \( 4^2 - 1^2 = 16 - 1 = 15 \), which is too much.
6. Changing one \( 1^2 \) to \( 3^2 \) increases the sum by \( 3^2 - 1^2 = 9 - 1 = 8 \). We still need to increase the sum by \( 13 - 8 = 5 \).
7. Changing another \( 1^2 \) to \( 3^2 \) increases the sum by another 8, which is again too much.
8. Changing one \( 1^2 \) to \( 2^2 \) increases the sum by \( 2^2 - 1^2 = 4 - 1 = 3 \). We still need to increase the sum by \( 5 - 3 = 2 \).
9. Changing another \( 1^2 \) to \( 2^2 \) increases the sum by another 3, which is again too much.
10. Therefore, no combination of changes from \( 1^2 \) to larger squares will result in the sum \( n^2 \) with \( n^2 - 13 \) elements.
11. Hence, \( S(n) \leq n^2 - 14 \).

### Part (b)
1. We need to find an integer \( n \) such that \( S(n) = n^2 - 14 \).
2. Consider \( n = 13 \).
3. We need to show that \( 13^2 = 169 \) can be written as the sum of \( 169 - 14 = 155 \) positive squares.
4. We can write \( 13^2 \) as follows:
   - \( 13^2 = 13^2 \)
   - \( 13^2 = 5^2 + 12^2 \)
   - \( 13^2 = 12^2 + 3^2 + 4^2 \)
   - \( 13^2 = 8^2 + 10^2 + 4^2 + 1^2 \)
   - \( 13^2 = 8^2 + 8^2 + 4^2 + 4^2 + 3^2 \)
5. Using the factorization \( 13^2 = 8^2 + 8^2 + 4^2 + 5^2 \), we get:
   - \( 8^2 = 64 \)
   - \( 4^2 = 16 \)
   - \( 5^2 = 25 \)
6. Summing these, we get \( 64 + 64 + 16 + 25 = 169 \).
7. Therefore, \( S(13) = 169 - 14 = 155 \).

### Part (c)
1. We need to prove that there are infinitely many integers \( n \) such that \( S(n) = n^2 - 14 \).
2. Assume \( n \) is such an integer. We need to show that \( 2n \) also works.
3. Consider \( 4n^2 \). We need to show that \( 4n^2 \) can be written as the sum of \( 4n^2 - 14 \) positive squares.
4. Since \( n^2 \) can be written as the sum of \( n^2 - 14 \) positive squares, we can write \( 4n^2 \) as the sum of \( 4(n^2 - 14) = 4n^2 - 56 \) positive squares.
5. We can use the same method to show that \( 4n^2 - 55 \), \( 4n^2 - 54 \), ..., \( 4n^2 - 14 \) can also be written as the sum of positive squares.
6. Therefore, if \( n \) works, then \( 2n \) also works.
7. By induction, there are infinitely many integers \( n \) such that \( S(n) = n^2 - 14 \).

The final answer is \( \boxed{ n = 13 } \) for part (b) and infinitely many integers for part (c).