Functional Equation: A, B, C Answers

[juantheron]

http://latex.codecogs.com/gif.latex?\hspace{-20}$%20A%20function%20$f:\mathbb{N}%20\rightarrow%20\mathbb{N}$%20and%20satisfies%20$f(ab)%20=%20f(a)+f(b)$.\\%20Where%20$a$%20and%20$b$%20are%20Coprime%20Natural%20no.\\%20and%20$f(c+d)%20=%20f(c)+f(d)\forall$%20prime%20no.%20$c$%20and%20$d$.%20Then\\%20(a)%20The%20value%20of%20$f(1)+f(2)+f(3)%20=%20$\\%20(b)%20$\frac{f(5)+f(7)}{f(4)}%20=$\\%20(c)%20$f(9)-f(6)+f(3)%20=%20$

 

[CaptainBlack]

http://latex.codecogs.com/gif.latex?\hspace{-20}$%20A%20function%20$f:\mathbb{N}%20\rightarrow%20\mathbb{N}$%20and%20satisfies%20$f(ab)%20=%20f(a)+f(b)$.\\%20Where%20$a$%20and%20$b$%20are%20Coprime%20Natural%20no.\\%20and%20$f(c+d)%20=%20f(c)+f(d)\forall$%20prime%20no.%20$c$%20and%20$d$.%20Then\\%20(a)%20The%20value%20of%20$f(1)+f(2)+f(3)%20=%20$\\%20(b)%20$\frac{f(5)+f(7)}{f(4)}%20=$\\%20(c)%20$f(9)-f(6)+f(3)%20=%20$

Since \(1\) is coprime to every natural \(f(1)=0\).

Also since for any prime \(c>2\) we have \(f(2c)=f(2)+f(c)\) and \(f(2c)=f(c+c)=f(c)+f(c)\) we conclude that \(f(c)=f(2)\), which is sufficient to allow us to answer (a), (b) and (c) in terms of \(f(2)\).

At present I don't see any means of evaluating \(f(2)\).CB

 

[Opalg]

Since \(1\) is coprime to every natural \(f(1)=0\).

Also since for any prime \(c>2\) we have \(f(2c)=f(2)+f(c)\) and \(f(2c)=f(c+c)=f(c)+f(c)\) we conclude that \(f(c)=f(2)\), which is sufficient to allow us to answer (a), (b) and (c) in terms of \(f(2)\).

CB

But that leads to something strange if you put $c=2$ and $d=3$, because it then follows that $f(2) = f(5) = f(2+3) = f(2) + f(3) = 2f(2).$ Thus $f(2)=0$ and hence $f(p)=0$ for every prime $p.$

 

[CaptainBlack]

But that leads to something strange if you put $c=2$ and $d=3$, because it then follows that $f(2) = f(5) = f(2+3) = f(2) + f(3) = 2f(2).$ Thus $f(2)=0$ and hence $f(p)=0$ for every prime $p.$

If that does not entail a contradiction then that gives us a full solution, including undefined for (b), it also answers the implied question I added to my post between you starting to reply and my seeing your reply :) . Alternativly there is a condition missing from the statement of the question.

CB