Let's find such numbers $a, b, c$ that for every positive integer $n$, the following equality holds:

$$
(n+3)^{2}=a \cdot(n+2)^{2}+b \cdot(n+1)^{2}+c \cdot n^{2}
$$
I. solution. Let's expand the parentheses on both sides:

$$
n^{2}+6 n+9=(a+b+c) n^{2}+(4 a+2 b) n+(4 a+b)
$$

The left and right sides of the equation will certainly be equal if the coefficients of the corresponding powers of $n$ are equal:

$$
\begin{aligned}
a+b+c & =1 \\
4 a+2 b & =6 \\
4 a+b & =9
\end{aligned}
$$

Subtracting equation (3) from equation (2), we get $b=-3$. From this, it easily follows that $a=3$ and $c=1$. Therefore, the equality always holds when $a=3, b=-3, c=1$.

Remark. Whether there are other suitable triples of numbers is not revealed by this solution, but it was not a question.

The solution can also be continued after rearrangement in a different way:

II. solution. If the values of two polynomials are equal at infinitely many points (in our case, on the positive integers), then they are equal for every real $n$. The values of two polynomials are equal for every real value of the variable if and only if the polynomials are equal coefficient by coefficient. From this, we similarly obtain that $a=3, b=-3$, and $c=1$. This triple of numbers (and only this) makes the two polynomials equal for every real, and thus for every positive integer $n$.

Remark. If the solver did not indicate whether they were working according to the I. or II. solution, we evaluated according to the II. method. Most 2-point solutions resulted from this lack.

III. solution. If the equality holds for all positive integers $n$ with the sought values of $a, b, c$, then it also holds for $n=1, n=2$, and $n=3$:

$$
\begin{aligned}
& 16=9 a+4 b+c \\
& 25=16 a+9 b+4 c \\
& 36=25 a+16 b+9 c
\end{aligned}
$$

Solving for $c$ from equation (1): $c=16-9 a-4 b$. Substituting this into (2) and solving for $a$: $a=-0.35 b+1.95$. Substituting this back into (1): $c=-1.55-0.85 b$. Substituting the expressions for $a$ and $c$ into (3) and rearranging the equation, we get $b=-3$. From this, $a=3$ and $c=1$.

We have found that if the problem has a solution, it can only be this triple of numbers. We must verify that we indeed obtained an identity that holds for every $n$. Substituting these values into the original equation:

$$
(n+3)^{2}=3(n+2)^{2}-3(n+1)^{2}+n^{2}
$$

Both sides of the equation are equal to the following expression: $n^{2}+6 n+9$, so it is indeed an identity.

Remark. 1. Many of those who followed the course of the III. solution forgot about the verification, they received 2 points.

2. The given formula provides a third-order recurrence for the sequence of squares of positive integers.