47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA

Solution:

Let $a = (2n+1)^5 - 2n - 1$.

First, expand $(2n+1)^5$ using the binomial theorem:

$$(2n+1)^5 = \sum_{k=0}^5 \binom{5}{k} (2n)^k (1)^{5-k} = 1 + 5 \cdot 2n + 10 \cdot (2n)^2 + 10 \cdot (2n)^3 + 5 \cdot (2n)^4 + (2n)^5$$

Calculate each term:

$1$

$5 \cdot 2n = 10n$

$10 \cdot (2n)^2 = 10 \cdot 4n^2 = 40n^2$

$10 \cdot (2n)^3 = 10 \cdot 8n^3 = 80n^3$

$5 \cdot (2n)^4 = 5 \cdot 16n^4 = 80n^4$

$(2n)^5 = 32n^5$

So,
$$(2n+1)^5 = 1 + 10n + 40n^2 + 80n^3 + 80n^4 + 32n^5$$

Therefore,
$$a = (2n+1)^5 - 2n - 1 = [1 + 10n + 40n^2 + 80n^3 + 80n^4 + 32n^5] - 2n - 1$$

$1 - 1 = 0$, $10n - 2n = 8n$

So,
$$a = 8n + 40n^2 + 80n^3 + 80n^4 + 32n^5$$

Factor $8n$:
$$a = 8n(1 + 5n + 10n^2 + 10n^3 + 4n^4)$$

Now, $240 = 2^4 \cdot 3 \cdot 5$.

We will show that $a$ is divisible by $16$, $3$, and $5$ for all positive integers $n$.

**Divisibility by $16$:**

Since $a = 8n(1 + 5n + 10n^2 + 10n^3 + 4n^4)$, $8n$ is always divisible by $8$. We need to check divisibility by $2$ more (to get $16$).

Consider $n$ even: $n = 2k$, then $8n$ is divisible by $16$.

If $n$ is odd: $n = 2k+1$, $8n$ is still divisible by $8$, but is it divisible by $16$?

But $1 + 5n + 10n^2 + 10n^3 + 4n^4$ is always even for integer $n$ (since $n$ odd, $n^2$ odd, $n^3$ odd, $n^4$ odd, so sum is odd + odd + even + even + even = odd, but times $8n$ which is even, so the product is divisible by $16$ for all $n$).

Alternatively, check $a$ modulo $16$:

$(2n+1)^5 \equiv (2n+1)^5 \pmod{16}$

But $2n+1$ is odd, so $2n+1 \equiv 1, 3, 5, 7, 9, 11, 13, 15 \pmod{16}$.

Compute $(2n+1)^5$ modulo $16$ for all odd residues:

But $x^5 \equiv x \pmod{16}$ for odd $x$ (since $x^4 \equiv 1 \pmod{16}$ for odd $x$), so $(2n+1)^5 \equiv 2n+1 \pmod{16}$.

Therefore,
$$a = (2n+1)^5 - 2n - 1 \equiv (2n+1) - 2n - 1 = 0 \pmod{16}$$

So $a$ is divisible by $16$.

**Divisibility by $3$:**

Compute $a \pmod{3}$:

$2n+1$ modulo $3$ can be $1$, $2$, or $0$.

Case 1: $2n+1 \equiv 0 \pmod{3}$, i.e., $2n \equiv -1 \pmod{3}$, $n \equiv 1 \pmod{3}$.

Then $a = 0^5 - 2n - 1 \equiv 0 - 2n - 1 \pmod{3}$
But $n \equiv 1$, so $a \equiv 0 - 2 \cdot 1 - 1 = -2 - 1 = -3 \equiv 0 \pmod{3}$

Case 2: $2n+1 \equiv 1 \pmod{3}$, $2n \equiv 0 \pmod{3}$, $n \equiv 0 \pmod{3}$
Then $a = 1^5 - 2n - 1 = 1 - 2n - 1 = -2n$
But $n \equiv 0$, so $a \equiv 0 \pmod{3}$

Case 3: $2n+1 \equiv 2 \pmod{3}$, $2n \equiv 1 \pmod{3}$, $n \equiv 2 \pmod{3}$
Then $a = 2^5 - 2n - 1 = 32 - 2n - 1 = 31 - 2n$
But $n \equiv 2$, so $2n \equiv 4 \equiv 1 \pmod{3}$
So $a \equiv 31 - 1 = 30 \equiv 0 \pmod{3}$

Therefore, $a$ is divisible by $3$ for all $n$.

**Divisibility by $5$:**

Compute $a \pmod{5}$:

$2n+1$ modulo $5$ can be $1, 2, 3, 4, 0$.

Case 1: $2n+1 \equiv 0 \pmod{5}$, $2n \equiv -1 \pmod{5}$, $n \equiv 2 \pmod{5}$
Then $a = 0^5 - 2n - 1 = 0 - 2n - 1$
But $n \equiv 2$, so $2n \equiv 4$, $a \equiv 0 - 4 - 1 = -5 \equiv 0 \pmod{5}$

Case 2: $2n+1 \equiv 1 \pmod{5}$, $2n \equiv 0$, $n \equiv 0$
$a = 1^5 - 2n - 1 = 1 - 0 - 1 = 0$

Case 3: $2n+1 \equiv 2$, $2n \equiv 1$, $n \equiv 3$
$a = 2^5 - 2n - 1 = 32 - 2n - 1 = 31 - 2n$
$n \equiv 3$, $2n \equiv 6 \equiv 1$, $a \equiv 31 - 1 = 30 \equiv 0 \pmod{5}$

Case 4: $2n+1 \equiv 3$, $2n \equiv 2$, $n \equiv 1$
$a = 3^5 - 2n - 1 = 243 - 2n - 1 = 242 - 2n$
$n \equiv 1$, $2n \equiv 2$, $a \equiv 242 - 2 = 240 \equiv 0 \pmod{5}$

Case 5: $2n+1 \equiv 4$, $2n \equiv 3$, $n \equiv 4$
$a = 4^5 - 2n - 1 = 1024 - 2n - 1 = 1023 - 2n$
$n \equiv 4$, $2n \equiv 8 \equiv 3$, $a \equiv 1023 - 3 = 1020 \equiv 0 \pmod{5}$

Therefore, $a$ is divisible by $5$ for all $n$.

Conclusion:

$a$ is divisible by $16$, $3$, and $5$, so $a$ is divisible by $\operatorname{lcm}(16,3,5) = 240$ for all positive integers $n$.