# Help on Mobius function

• Apr 16th 2008, 12:41 AM
kleenex
Help on Mobius function
Show that if $n$ is a positive integer, then
$\mu(n)\mu(n+1)\mu(n+2)\mu(n+3) = 0$

I know that's mean one of the $\mu$ above has a square factor.
Sure at least one $\mu(n),\mu(n+1),\mu(n+2),\mu(n+3)$ of it is even and it must be power of 2 but how can I write it down properly as a proof?

• Apr 16th 2008, 03:54 AM
PaulRS
Among 4 consecutive natural numbers one must be divisble by 4 (Wink)

You can separate in cases $n\equiv{0;1;2;3}(\bmod.4)$