In Andreescu's Number Theory: Structures, Examples, and Problems, the author offers a problem (2.1.19), and its respective, from the 33rd IMO:
Problem 2.1.19.
For each positive integer n, denote by s(n) the greatest integer such that for all positive integers k ≤ s(n), $n^2 $ can be expressed as a sum of squares of k positive integers.
(a) Prove that $s(n) ≤ n^2 − 14 $ for all n ≥ 4.
(b) Find a number n such that $s(n) = n^2 − 14 $.
(c) Prove that there exist infinitely many positive integers n such that $s(n) = n^2 − 14 $.
Solution.
(a) Representing $n^2 $as a sum of $n^2 − 13 $ squares is equivalent to representing 13 as a sum of numbers of the form $x^2 −1 $, x ∈ N, such as 0, 3, 8, 15,... But it is easy to check that this is impossible, and hence $s(n) ≤ n^2 − 14 $.
(b) Let us prove that $s(13) = 13^2 − 14 = 155 $. Observe that $13^2 = 8^2 + 8^2 + 4^2 + 4^2 + 3^2 $ $= 8^2 + 8^2 + 4^2 + 4^2 + 2^2 + 2^2 + 1^2 $ $= 8^2 + 8^2 + 4^2 + 3^2 + 3^2 + 2^2 + 1^2 + 1^2 + 1^2 $. Given any representation of $n^2 $ as a sum of m squares one of which is even, we can construct a representation as a sum of m + 3 squares by dividing the square into four equal squares. Thus the first equality enables us to construct representations with 5, 8, 11,..., 155 squares, the second to construct ones with 7, 10, 13,..., 154 squares, and the third with 9, 12,..., 153 squares. It remains only to represent $13^2 $ as a sum of k = 2, 3, 4, 6 squares. This can be done as follows: $13^2 = 12^2 + 5^2 = 12^2 + 4^2 + 3^2 $ $= 11^2 + 4^2 + 4^2 + 4^2 $ $= 12^2 + 3^2 + 2^2 + 2^2 + 2^2 + 2^2 $.
(c) We shall prove that whenever $s(n) = n^2 − 14 $ for some n ≥ 13, it also holds that $s(2n) = (2n)^2 − 14 $. This will imply that $s(n) = n^2 − 14 $ for any $n = 2^t · 13 $. If $n^2 = x_1^2 +···+ x_r^2 $, then we have $(2n)^2 = (2x_1)^2 +···+(2x_r)^2 $. Replacing $(2x_i)^2 $ with $x_i^2 + x_i^2 +x_i^2 +x_i^2 $ as long as it is possible, we can obtain representations of $(2n)^2 $ consisting of $r,r +3,..., 4r $ squares. This gives representations of $(2n)^2 $ into k squares for any k ≤ $4n^2−62 $. Further, we observe that each number m ≥ 14 can be written as a sum of k ≥ m numbers of the form $x^2 − 1 $, x ∈ N, which is easy to verify. Therefore if $2n^2 ≤ k ≤ 4n^2 − 14 $, it follows that $4n^2 − k $ is a sum of k numbers of the form $x^2 − 1 $ (since k ≥ $4n^2 − k $≥ 14), and consequently $4n^2 $ is a sum of k squares.
I cannot solve the remark Andreescu shared for this problem.
Remark. One can find exactly the value of $s(n) $ for each n:
$s(n)= $
$=1 $, if n has a prime divisor congruent to 3 mod 4
$=2 $, if n is of form $5\cdot2^k $, where k is a positive integer
$=n^2-14 $, otherwise.