Is f = n^2 a completely multiplicative function?

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Discussion Overview

The discussion revolves around the nature of the function f = n^2 and whether it qualifies as a completely multiplicative function. Participants explore the implications of this classification, particularly in relation to the sum of divisors function F and its properties.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that if f is completely multiplicative, then the sum \sum_{d \mid n} f(d) being completely multiplicative is not necessarily true, indicating a potential counterexample exists.
  • Another participant introduces the sum F and considers the expression F(2)F(2) - F(4) in relation to the discussion.
  • A participant questions whether F(4) - F(2)F(2) equals zero, expressing a desire for a more explicit function f that demonstrates F is not completely multiplicative.
  • Another participant proposes that the issue may arise from using two arguments with a common factor, suggesting an example with different primes (5 and 7) to illustrate the point.
  • One participant reiterates the desire for an explicit function f, specifically stating f = n^2.

Areas of Agreement / Disagreement

Participants express differing views on the properties of the function f and its implications for the sum F. No consensus is reached regarding whether f = n^2 is completely multiplicative or not, and the discussion remains unresolved.

Contextual Notes

Participants note that the behavior of the function may depend on the specific properties of the arguments used in the sums, particularly regarding common factors and the nature of the primes involved.

math_grl
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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.
 
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call the sum F
consider
F(2)F(2)-F(4)
 
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.
 
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!
 
math_grl said:
I was hoping for more an explicit function, f, such that F is not completely multiplicative.

let f=n^2
 

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