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Number Theory functional equation

jacks

Well-known member
Apr 5, 2012
226
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
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
Jan 26, 2012
890
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