Is f = n^2 a completely multiplicative function?

[math_grl]

If $$f$$ is completely multiplicative, then $$\sum_{d \mid n} f(d)$$ is completely multiplicative is not true. There must be an easy counterexample for this yet I cannot come up with one.

 

[lurflurf]

call the sum F
consider
F(2)F(2)-F(4)

 

[math_grl]

So you are saying that F(4) - F(2)F(2) is not 0 as it should be?
I was hoping for more an explicit function, f, such that F is not completely multiplicative.

 

[dodo]

What lurflurf said should work with any completely multiplicative f... I believe the key issue is that the two arguments, 2 and 2, have a factor in common.

For example, try what happens with two different primes, say 5 and 7:
F(5)F(7) = (f(1) + f(5)) . (f(1) + f(7))
and
F(35) = f(1) + f(5) + f(7) + f(35)
When you distribute the parenthesis in the first equation, and apply f(a) . f(b) = f(ab), you should get the same RHS as the second equation.

Now try with the values lurflurf proposed, and see what happens!

 

[lurflurf]

I was hoping for more an explicit function, f, such that F is not completely multiplicative.

let f=n^2