the Mobius function

[Emeric Deutsch]

Dear Seqfans,
It is well-known that for n>1 we have  

   SUM_{d|n}mobius(d) = 0.

Is it known or easy to see that for n>2 we have 

  SUM_{d|n}(-1)^(n/d)*mobius(d) = 0 ?

Thanks.
Emeric