Why is this arbitrary-looking identity of arithmetic functions "obvious"?The question asks us the following: if f is a multiplicative arithmetic function, and g is a completely multiplicative arithmetic function, and furthermore for all primes p and n ≥ 1 we have the relation f ( p n + 1 ) = f ( p ) f ( p n ) − g ( p ) f ( p n − 1 ) ,then for all n,m f ( n ) f ( m ) = ∑ d ∣ gcd ( n , m ) g ( d ) f ( n m d 2 ) .More than any exercise so far in this book, this looks completely arbitrary to me. I have solved it: you can consider the question only for n = p a , m = p b , and then it follows for arbitrary n,m by multiplicativity of f,g; and we can prove it for p a ≤ p b by induction on a. But that has really taught me nothing: I don't understand why it might make some sense that this is true.If g ( p ) = 0 for all primes p, then the relation we start from becomes f ( p n ) = f ( p ) n - i.e., f is completely multiplicative. Thus maybe we can see the g ( p ) f ( p n − 1 ) as a sort of error term of the complete multiplicativity of f. But this doesn't help me see why the result makes sense.Another approach I tried was noting that the relation allows us to write f ( p k ) as a polynomial in f(p),g(p), but I wrote out a few and I didn't see a pattern I understood.So my question is: where does this identity come from? How should I understand it, visualize it, "get" it? How would one come up with an exercise like this?

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Why is this arbitrary-looking identity of arithmetic functions &quot;obvious&quot;?The question asks us the following: if f is a multiplicative arithmetic function, and g is a completely multiplicative arithmetic function, and furthermore for all primes p and   n  ≥  1 we have the relation  f  (      p          n      +      1        )  =  f  (  p  )  f  (      p    n    )  −  g  (  p  )  f  (      p          n      −      1        )  ,then for all n,m  f  (  n  )  f  (  m  )  =      ∑          d      ∣      gcd      (      n      ,      m      )        g  (  d  )  f      (                  n        m                    d        2              )    .More than any exercise so far in this book, this looks completely arbitrary to me. I have solved it: you can consider the question only for   n  =      p    a    ,  m  =      p    b  , and then it follows for arbitrary n,m by multiplicativity of f,g; and we can prove it for       p    a    ≤      p    b   by induction on a. But that has really taught me nothing: I don&#039;t understand why it might make some sense that this is true.If   g  (  p  )  =  0 for all primes p, then the relation we start from becomes   f  (      p    n    )  =  f  (  p      )    n   - i.e., f is completely multiplicative. Thus maybe we can see the   g  (  p  )  f  (      p          n      −      1        ) as a sort of error term of the complete multiplicativity of f. But this doesn&#039;t help me see why the result makes sense.Another approach I tried was noting that the relation allows us to write   f  (      p    k    ) as a polynomial in f(p),g(p), but I wrote out a few and I didn&#039;t see a pattern I understood.So my question is: where does this identity come from? How should I understand it, visualize it, &quot;get&quot; it? How would one come up with an exercise like this?

DoryErrofbi · Accepted Answer

Step1 This identity a generalisation of the fact that  f  (  m  )  f  (  n  )  =  f  (  m  n  )whenever   (  m  ,  n  )  =  1, which holds for any multiplicative function f. On seeing this, it&#039;s a natural question what happens when m and n are not coprime.So we&#039;re asking for what f(m)f(n) is for a multiplicative function for general n. We expect that f(mn) is a &#039;main term&#039; - indeed, if f is completely multiplicative, then this is exactly what it is. And we know that it is also what it is when   (  m  ,  n  )  =  1. So we expect something like$$   f  (  m  )  f  (  n  )  =  f  (  m  n  )  +  E$where E is some correction term that should depend only on (m,n) and mn, and should vanish if   (  m  ,  n  )  =  1 or if f is completely multiplicative.Step 2The exercise gives an explicit form for E. As you say, there is some amount of calculation involved, but a small amount of trial and error would naturally lead to the correct form. One might begin, for example, by saying that we should measure how far the multiplicative f is from being completely multiplicative. Completely multiplicative means   f  (      p          n      +      1        )  =  f  (  p  )  f  (      p    n    ), so one might define the function   F  (      p          n        )  =  f  (      p          n      +      1        )  −  f  (  p  )  f  (      p    n    ).If you try and put that into   f  (  m  )  f  (  n  )  −  f  (  m  n  )  −  E then it becomes apparent that it would be very convenient for an inductive argument if   F  (      p    n    ) could be &#039;factored&#039; as   g  (  p  )  f  (      p          n      −      1        ). Thus the hypothesis in the exercise.

Iris Vaughn · Accepted Answer

Step 1There is nothing more than what you said : it is an equation between operators on multiplicative functions so we look at   n  =      p    a    ,  m  =      p    b   where the equation takes a much simpler form. It comes from modular forms and Euler products of the form       ∏    p        1          1      −      f      (      p      )              p                  −          s                    +      g      (      p      )              p                  −          2          s                      =      ∑    n    f  (  n  )      n          −      s      Step 2The eigenforms (modular forms with multiplicative coefficients) satisfy those relations with   g  (  p  )  =      p          k      −      1       or   g  (  p  )  =  χ  (  p  )      p          k      −      1      . Modular forms are the topics of other books by Apostol which explains why it appears as an exercice in Introduction to Analytic Number Theory.

Why is this arbitrary-looking identity of arithmetic functions "obvious"? If f is a multiplicative arithmetic function, and g is a completely multiplicative arithmetic function, and furthermore for all primes p and n>=1 we have the relation f(p^{n+1})=f(p)f(p^n)-g(p)f(p^{n-1})

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