
Help on Mobius function
Show that if $\displaystyle n$ is a positive integer, then
$\displaystyle \mu(n)\mu(n+1)\mu(n+2)\mu(n+3) = 0$
I know that's mean one of the $\displaystyle \mu$ above has a square factor.
Sure at least one $\displaystyle \mu(n),\mu(n+1),\mu(n+2),\mu(n+3)$ of it is even and it must be power of 2 but how can I write it down properly as a proof?
Help you for your help.

Among 4 consecutive natural numbers one must be divisble by 4 (Wink)
You can separate in cases $\displaystyle n\equiv{0;1;2;3}(\bmod.4)$