2. Let M be the set of all values of the expression $V(n)=n^{4}+11 n^{2}-12$, where $n$ is an odd natural number. Find all possible remainders when dividing by 48 that the elements of set M can give.
2. First, we calculate the relevant values of the expression $V$ for the first few odd numbers:

$$
\begin{aligned}
& \begin{array}{lc}
n & V(n) \\
\hline 1 & 0
\end{array} \\
& 3 \quad 168=3 \cdot 48+24 \\
& 5 \quad 888=18 \cdot 48+24 \\
& 7 \quad 2928=61 \cdot 48 \\
& 9 \quad 7440=155 \cdot 48
\end{aligned}
$$

Thus, the remainders we are looking for include the numbers 0 and 24. We will show that no other remainders are possible. To do this, it suffices to prove that for every odd number $n$, $24 \mid V(n)$. From the homework round, we know that for every natural number $n$, $12 \mid V(n)$, and thus $3 \mid V(n)$. Since the numbers 3 and 8 are coprime, it suffices to show that for every odd number $n$, $8 \mid V(n)$. We will use the factorization of the given expression:

$$
V(n)=n^{4}+11 n^{2}-12=\left(n^{2}-1\right)\left(n^{2}+12\right)=(n-1)(n+1)\left(n^{2}+12\right).
$$

Any odd natural number $n$ can be written in the form $n=2 k-1$, where $k \in \mathbb{N}$. For such $n$, we then get

$$
V(2 k-1)=[(2 k-1)-1][(2 k-1)+1]\left[(2 k-1)^{2}+12\right]=4(k-1) k\left(4 k^{2}-4 k+13\right),
$$

and since the product $(k-1) k$ of two consecutive integers is divisible by two, the entire expression is divisible by eight.

Conclusion. The given expression can yield remainders of 0 and 24 when divided by 48.

Note. The fact that $8 \mid V(n)$ for every odd $n$ can also be proven differently, without using the factorization (1). If $n=2 k-1$, where $k \in \mathbb{N}$, then the number

$$
n^{2}=(2 k-1)^{2}=4 k^{2}-4 k+1=4 k(k-1)+1
$$

gives a remainder of 1 when divided by eight (since one of the numbers $k, k-1$ is even), and thus the number $n^{4}$ (as the square of the odd number $n^{2}$) also gives a remainder of 1. Therefore, $n^{2}=8 u+1$ and $n^{4}=8 v+1$ for suitable integers $u$ and $v$, so the value of the expression

$$
V(2 k-1)=(8 v+1)+11(8 u+1)-12=8(v+11 u)
$$

is indeed a multiple of eight.

Let's also include a similar proof of the fact $3 \mid V(n)$ from the homework round. For numbers $n$ divisible by three, this is obvious; other $n$ are of the form $n=3 k \pm 1$, so the number

$$
n^{2}=(3 k \pm 1)^{2}=9 k^{2} \pm 6 k+1=3 k(3 k \pm 2)+1
$$

gives a remainder of 1 when divided by three, and the same is true for the number $n^{4}=\left(n^{2}\right)^{2}$. Substituting $n^{2}=3 u+1$ and $n^{4}=3 v+1$ into the expression $V(n)$ directly leads to the conclusion that $3 \mid V(n)$.

For a complete solution, award 6 points. For using the fact from the homework round ($12 \mid V(n)$ for any $n$) to deduce (or state) that the remainders can only be $0, 12, 24$, and 36, award 1 point. For an attempt to prove that $8 \mid V(n)$ for odd $n$, award 1 point; for a proof of this statement, award 2 points. For the conclusion that $3 \mid V(n)$ and $8 \mid V(n)$ imply $24 \mid V(n)$, award 1 point (the coprimality of the numbers 3 and 8 must be mentioned; a conclusion like "from $12 \mid V(n)$ and $8 \mid V(n)$ it follows that $24 \mid V(n)$" is insufficient if, for example, it is not added that 24 is the least common multiple of 8 and 12). For verifying that the remainders 0 and 24 are indeed possible, award 1 point (it suffices to mention, for example, the values $V(1)$ and $V(3)$).