# Number Theoryfunctional equation

#### CaptainBlack

##### Well-known member
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

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#### Opalg

##### MHB Oldtimer
Staff member
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

##### Well-known member
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