技术问询：证明对所有整数n，2n³+3n²+n可被2、3及6整除

Alright, let's work through these two proof tasks together. I'll break each part down with clear steps so you can follow along easily.

Task 1: Prove $2n^3 + 3n^2 + n$ is divisible by both 2 and 3 for all $n ∈ ℤ$ #

First, let's factor the expression to simplify things—factoring is usually a solid first step for divisibility proofs:
$$2n^3 + 3n^2 + n = n(2n^2 + 3n + 1) = n(n+1)(2n+1)$$

Prove divisibility by 2 #

Notice that $n$ and $n+1$ are consecutive integers. In any pair of consecutive integers, one must be even (divisible by 2). Since our factored expression includes both $n$ and $n+1$, the entire product will always have an even factor—so it's divisible by 2 for any integer $n$.

If you prefer a case-based approach:

• If $n$ is even ($n=2k$ for some integer $k$), the first factor $n$ is even, making the whole product even.
• If $n$ is odd, then $n+1$ is even, so again the product is even.

Prove divisibility by 3 #

We can use modular arithmetic here, testing all possible residues of $n$ modulo 3 (any integer falls into one of three categories: 0, 1, or 2 when divided by 3):

• Case 1: $n ≡ 0 \pmod{3}$
  Substitute into the original expression: $2(0)^3 + 3(0)^2 + 0 = 0$, which is obviously divisible by 3.
• Case 2: $n ≡ 1 \pmod{3}$
  Substitute: $2(1)^3 + 3(1)^2 + 1 = 2 + 3 + 1 = 6$, and $6$ is divisible by 3.
• Case 3: $n ≡ 2 \pmod{3}$
  Substitute: $2(2)^3 + 3(2)^2 + 2 = 16 + 12 + 2 = 30$, and $30$ is divisible by 3.

No matter what integer $n$ we pick, the expression is congruent to 0 modulo 3—so it's divisible by 3.

Task 2: Prove the expression is divisible by 6 #

You already noted this is equivalent to proving divisibility by both 2 and 3, which we just confirmed in Task 1. Here's why that equivalence holds:

• 2 and 3 are coprime integers (their greatest common divisor is 1). By a basic number theory rule, if a number is divisible by two coprime integers, it's divisible by their product. Since $2×3=6$, our expression—being divisible by both 2 and 3—must be divisible by 6.

That's all there is to it! Both proofs rely on breaking the problem into small, manageable parts, either through factoring or testing cases with modular arithmetic.

内容的提问来源于stack exchange，提问作者student