We seek the largest integer that divides $P$ for all odd $n$. Since $n$ is odd, all terms are odd, so $P$ is odd — 2 does not divide it.

February 22, 2026

We seek the largest integer that divides $P$ for all odd $n$. Since $n$ is odd, all terms are odd, so $P$ is odd — 2 does not divide it.

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Question: What is the largest integer that must divide the product of any four consecutive odd integers?
Solution: Let the four consecutive odd integers be $n, n+2, n+4, n+6$, where $n$ is odd. Their product is:
P = n(n+2)(n+4)(n+6)
We analyze divisibility by small primes:
Divisibility by 3**: Among any four consecutive odd integers, at least one is divisible by 3. This can be shown by checking modulo 3: the residues of odd integers mod 3 are 0, 1, or 2. In any block of 6 consecutive integers, four odd ones must include one divisible by 3. So $3 \mid P$.
Divisibility by 5**: Not guaranteed — e.g., $n = 1$: $1 \cdot 3 \cdot 5 \cdot 7 = 105$, divisible by 5, but $n = 7$: $7 \cdot 9 \cdot 11 \cdot 13 = 9009$, not divisible by 5. So 5 does not always divide $P$.

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