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Let a belong to the positive integers, and let p belong to the prime numbers, p > 3.
Can you prove
a(a + p)(a + 2p) is divisible by 6?
Let p be a prime with p>3, a a positive integer (actually any integer) and n=a(a+p)(a+2p). Then n is divisible by 6.
First if a is even, then n is divisible by 2; if a is odd, a+p is even and again n is divisible by 2. So n is divisible by 2.
Since p>3, p is not congruent to 0 mod 3. So p is 1 or 2 mod 3. Hence mod 3, n is congruent to a(a+1)(a+2) mod 3 or n is congruent to a(a+2)(a+1) mod 3. That is n is a(a+1)(a+2) mod 3. If a is 0, mod 3, 3 divides n. If a is 1 mod 3, a+2 is 0 mod 3 and again 3 divides n. If a is 2 mod 3, a+1 is 0 mod 3 and again 3 divides n.
Since 2 divides n and 3 divides n, 6 divides n.
I would take a similar approach, but use the fact that $\displaystyle p \equiv \pm 1 \pmod{6}$ as the starting point rather than $\displaystyle p \equiv \pm 1 \pmod{3}$. This probably removes all worries about $\displaystyle a$ being odd or even although I haven't worked it through. It does add some complication to the evaluation of $\displaystyle (a+p) \pmod{6}$ and $\displaystyle (a+2p) \pmod{6}$ though.