(a) Determine $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ such that the equality

$$
(n+2)^{2}=a(n+1)^{2}+b n^{2}+c(n-1)^{2}
$$

is true for any number $n$.

(b) Suppose that $x_{1}, x_{2}, \ldots, x_{7}$ satisfy the system

$$
\left\{\begin{array}{l}
x_{1}+4 x_{2}+9 x_{3}+16 x_{4}+25 x_{5}+36 x_{6}+49 x_{7}=1 \\
4 x_{1}+9 x_{2}+16 x_{3}+25 x_{4}+36 x_{5}+49 x_{6}+64 x_{7}=12 \\
9 x_{1}+16 x_{2}+25 x_{3}+36 x_{4}+49 x_{5}+64 x_{6}+81 x_{7}=123
\end{array}\right.
$$

Determine the value of

$$
16 x_{1}+25 x_{2}+36 x_{3}+49 x_{4}+64 x_{5}+81 x_{6}+100 x_{7}
$$
(a) If a polynomial vanishes for infinitely many values, then all its coefficients are zero.

Expanding the equality we have

$$
n^{2}+4 n+4=a\left(n^{2}+2 n+1\right)+b n^{2}+c\left(n^{2}-2 n+1\right)
$$

Thus,

$$
(a+b+c-1) n^{2}+(2 a-2 c-4) n+(a+c-4)=0
$$

for any number $n$. Therefore,

$$
\left\{\begin{array}{l}
a+b+c-1=0 \\
2 a-2 c-4=0 \\
a+c-4=0
\end{array}\right.
$$

Solving the system, we find $a=3, c=1$ and $b=-3$.

(b) Let

$$
\begin{aligned}
& S_{1}=x_{1}+4 x_{2}+9 x_{3}+16 x_{4}+25 x_{5}+36 x_{6}+49 x_{7}=1 \\
& S_{2}=4 x_{1}+9 x_{2}+16 x_{3}+25 x_{4}+36 x_{5}+49 x_{6}+64 x_{7}=12 \\
& S_{3}=9 x_{1}+16 x_{2}+25 x_{3}+36 x_{4}+49 x_{5}+64 x_{6}+81 x_{7}=123
\end{aligned}
$$

By the identity from part (a), we have that

$$
\begin{aligned}
& 16 x_{1}+25 x_{2}+36 x_{3}+49 x_{4}+64 x_{5}+81 x_{6}+100 x_{7}= \\
& \quad=3 S_{3}-3 S_{2}+S_{1}=3 \cdot 123-3 \cdot 12+1=334
\end{aligned}
$$

Hint: Write the number as $10 a+b$, where $b$ is a digit.
Hint: Let $S_{n}$ be the sum of such divisors, calculate the difference $S_{n}-$ $S_{n-1}$.

Helpful Facts: The sum of the first $n$ odd numbers is

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