My question is well written in title. I thought that something like divisibility by $3$ will work but it is not. Please help.
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7Are you sure divisibility by 3 doesn't work? – Aka_aka_aka_ak Jan 08 '17 at 18:41
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2$n$ is one of $0,1,2 \pmod 3$. Just try each case. – lulu Jan 08 '17 at 18:42
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Possibly related: http://math.stackexchange.com/questions/64414/show-that-every-prime-p3-is-either-of-the-form-6n1-or-of-the-form-6n5 – GDumphart Jan 08 '17 at 18:46
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Let $m=n+2$; then the product of the three numbers is $$\begin{array}{rcl}(m-2)m(m+2)&=&m^3-4m \\ &\equiv& m^3-m\pmod3 \\ &\equiv& 0\pmod3\end{array}$$ since $m^3\equiv m\pmod3$. So $3$ divides the product of the numbers and so it divides one of them. As $n>3$, that number cannot be $3$ itself; hence it is not prime.
George Law
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Product of three consecutive integers is divisible by $3!.$ I just want to mention it. – Bumblebee Jan 08 '17 at 19:31
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if $$n\equiv 1\mod 3$$ then we have $$n+2\equiv 0 \mod 3$$ if $$n\equiv 2 \mod 3$$ then we have $$n+4\equiv 0 \mod 3$$ note that if we have $$n=3$$ then we get the prime numbers $$3,5,7$$
Dr. Sonnhard Graubner
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Hint: Notice that when $n\ge4$, $$3|n(n+2)(n+4)$$ If you want to prove it, consider making cases when $n$ is even and when $n$ is odd.
Vidyanshu Mishra
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