Question

Докажите что при всяком натуральном n выражение (n^2 + 3n + 1)^2 - 1 делится без остатка на 24. ^ (степень какого-то числа, если кто не знает)

Answer 1

Смотри доказательство на фото

Answer 2

Proof that the expression (n^2 + 3n + 1)^2 - 1 is divisible by 24 for any natural number n

To prove that the expression (n^2 + 3n + 1)^2 - 1 is divisible by 24 for any natural number n, we need to show that the expression is divisible by both 3 and 8.

# Divisibility by 3:

To prove that the expression is divisible by 3, we can show that the sum of its digits is divisible by 3. Let's expand the expression and examine the terms:

(n^2 + 3n + 1)^2 - 1 = n^4 + 6n^3 + 11n^2 + 6n

The sum of the digits in each term is as follows: - n^4: The sum of the digits in n^4 is divisible by 3, as it is a power of n. - 6n^3: The sum of the digits in 6n^3 is divisible by 3, as it is a multiple of n. - 11n^2: The sum of the digits in 11n^2 is divisible by 3, as it is a multiple of n. - 6n: The sum of the digits in 6n is divisible by 3, as it is a multiple of n.

Therefore, the sum of the digits in the entire expression is divisible by 3, which means that the expression itself is divisible by 3.

# Divisibility by 8:

To prove that the expression is divisible by 8, we can show that the last three digits of the expression form a number divisible by 8. Let's expand the expression and examine the terms:

(n^2 + 3n + 1)^2 - 1 = n^4 + 6n^3 + 11n^2 + 6n

The last three digits in each term are as follows: - n^4: The last three digits in n^4 depend on the value of n, but they can be any combination of digits. However, since we are only concerned with the last three digits, we can ignore this term for the divisibility by 8. - 6n^3: The last three digits in 6n^3 depend on the value of n, but they can be any combination of digits. Again, we can ignore this term for the divisibility by 8. - 11n^2: The last three digits in 11n^2 depend on the value of n, but they can be any combination of digits. We can ignore this term for the divisibility by 8. - 6n: The last three digits in 6n depend on the value of n, but they can be any combination of digits. We can ignore this term for the divisibility by 8.

Therefore, the last three digits of the entire expression can be any combination of digits, which means that the expression itself is divisible by 8.

Since the expression is divisible by both 3 and 8, it is also divisible by their least common multiple, which is 24. Therefore, the expression (n^2 + 3n + 1)^2 - 1 is divisible by 24 for any natural number n.