2. Find the greatest natural number $d$ that has the property that for any natural number $n$, the value of the expression

$$
V(n)=n^{4}+11 n^{2}-12
$$

is divisible by the number $d$.
SOLUTION. First, let's calculate the values of $V(n)$ for the smallest natural numbers $n$ and write their prime factorizations in a table:

| $n$ | 1 | 2 | 3 | 4 |
| :---: | :---: | :---: | :---: | :---: |
| $V(n)$ | 0 | $48=2^{4} \cdot 3$ | $168=2^{3} \cdot 3 \cdot 7$ | $420=2^{2} \cdot 3 \cdot 5 \cdot 7$ |

From this, we see that the sought divisor $d$ of all numbers $V(n)$ must be a divisor of the number $2^{2} \cdot 3=12$, satisfying the inequality $d \leqq 12 .{ }^{1}$ If we can show that the number $d=12$ meets the conditions, i.e., that $V(n)$ is a multiple of 12 for every natural $n$, we will be done with the solution.

Transformation

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

where we have "factored out" the binomial $12 n^{2}-12$ from the expression $V(n)$, which is clearly a multiple of 12, reducing our task to verifying the divisibility of the binomial $n^{4}-n^{2}$ by 12 (i.e., by 3 and 4). For this, we use its factorization

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

For any integer $n$, the expression $n^{4}-n^{2}$ is certainly divisible by three (since one of the three consecutive integers $n-1, n, n+1$ is divisible by three) and simultaneously by four (which is guaranteed by the factor $n^{2}$ in the case of even $n$, and by the two even factors $n-1$ and $n+1$ in the case of odd $n$).

Let's add that the divisibility of the expression $V(n)$ by 12 can also be proven in other ways, for example, by using the factorization

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

or by working with the binomial $n^{4}+11 n^{2}$ and similarly.

Answer. The sought number $d$ is equal to 12.

## GUIDING AND SUPPLEMENTARY PROBLEMS:

N1. Prove that in the infinite sequence of numbers

$$
1 \cdot 2 \cdot 3,2 \cdot 3 \cdot 4,3 \cdot 4 \cdot 5,4 \cdot 5 \cdot 6, \ldots
$$

the first number is a divisor of all subsequent numbers. [Use the fact that among two, or three consecutive numbers, there is always a number divisible by two, or three, respectively.]

N2. Find all integers $d>1$ such that the values of the expressions $U(n)=n^{3}+17 n^{2}-1$ and $V(n)=$ $=n^{3}+4 n^{2}+12$ give the same remainders when divided by $d$, regardless of the integer $n$ chosen. [Only $d=13$ satisfies. The sought $d$ are precisely those that divide the difference $U(n)-V(n)=13 n^{2}-13=13(n-1)(n+1)$ for every integer $n$. To show that (apparently suitable) $d=13$ is the only one, substitute the value $n=d$ into the difference $U(n)-V(n)$: the number $d$ is coprime with the numbers $d-1$ and $d+1$, so it divides the product $13(d-1)(d+1)$ only if it divides the factor 13, i.e., when $d=13$.]

D1. For which natural numbers $n$ is the expression $V(n)=n^{4}+11 n^{2}-12$ not a multiple of eight? [The expression $V(n)=(n-1)(n+1)\left(n^{2}+12\right)$ is certainly a multiple of eight in the case of odd $n$, because $n-1$ and $n+1$ are two consecutive even numbers, so one of them is divisible by four, and the product of both is thus a multiple of eight. Since the product $(n-1)(n+1)$ is odd for even $n$, we are looking for those $n$ of the form $n=2 k$ for which the expression $n^{2}+12=4\left(k^{2}+3\right)$ is not divisible by eight, which occurs precisely when $k$ is even. The sought $n$ are therefore precisely those that are divisible by four.]

D2. Prove that for any integers $n$ and $k$ greater than 1, the number $n^{k+2}-n^{k}$ is divisible by twelve. $[59-\mathrm{C}-\mathrm{II}-1]$

D3. Prove that the expressions $23 x+y$ and $19 x+3 y$ are divisible by 50 for the same pairs of natural numbers $x$ and $y$. $[60-\mathrm{C}-\mathrm{I}-2]$

D4. Determine all integers $n$ for which $2 n^{3}-3 n^{2}+n+3$ is a prime number. $[62-C-I-5]$

D5. Prove that for every odd natural number $n$, the sum $n^{4}+2 n^{2}+2013$ is divisible by 96. $[63-\mathrm{C}-\mathrm{I}-5]$