正整数除法证明求助：求证对任意整数n≥1，6|n(n+1)(n+2)

云原生阿岚

2026-5-19

Proof that $6|n(n + 1)(n + 2)$ for all integers $n \geq 1$

Hey there! Let's work through this proof together—it's actually simpler once you break down what divisibility by 6 means. Since 6 is the product of two coprime integers (2 and 3, meaning their only common divisor is 1), we just need to show two things:

$n(n+1)(n+2)$ is divisible by 2
$n(n+1)(n+2)$ is divisible by 3

If both are true, then by the fundamental theorem of arithmetic, the product must be divisible by $2 \times 3 = 6$.

Step 1: Prove divisibility by 2

Consider any three consecutive integers: $n$, $n+1$, $n+2$. In any pair of consecutive integers, one must be even (divisible by 2). Since we have three in a row, there's guaranteed to be at least one even number in the set. For example:

If $n$ is even: $n$ itself is divisible by 2
If $n$ is odd: $n+1$ will be even, so it's divisible by 2

Either way, the product $n(n+1)(n+2)$ includes a factor of 2, so it's divisible by 2.

Step 2: Prove divisibility by 3

Similarly, in any three consecutive integers, one must be divisible by 3. This is because when you divide any integer by 3, the possible remainders are 0, 1, and 2. Let's cover all cases:

If $n \equiv 0 \pmod{3}$: $n$ is directly divisible by 3
If $n \equiv 1 \pmod{3}$: $n+2 = 1+2 = 3 \equiv 0 \pmod{3}$, so $n+2$ is divisible by 3
If $n \equiv 2 \pmod{3}$: $n+1 = 2+1 = 3 \equiv 0 \pmod{3}$, so $n+1$ is divisible by 3

No matter what value $n$ takes, one of the three terms is a multiple of 3, so the product includes a factor of 3.

Putting it all together

Since $n(n+1)(n+2)$ is divisible by both 2 and 3, and 2 and 3 are coprime, the product must be divisible by their least common multiple—which is exactly 6.

Let's test with a few concrete values to confirm: