I'm interesting to know more about multiplicative property of integer functions then I'd like to ask this humble question:

Question: Is it possible to find all integer functions which satisfy $f(m!+n!)\mid f(m!)+f(n!)$ and $m+n \mid f(m)+f(n)$?

Note: the symbol | meant divides

Thank you for any help

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    $\begingroup$ Can you give any context why you are interested in this question? In the present form the question looks suspiciously like a competition problem. $\endgroup$ – Jan-Christoph Schlage-Puchta Oct 29 '16 at 17:57
  • $\begingroup$ ok , really i 'd like to study and know more about multiplicative integer function and to study it's periodicity $\endgroup$ – ßłặck Pěặřł Oct 29 '16 at 18:02
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    $\begingroup$ It is tempting to say that this looks like an attempt to tempt people into solving a contest problem for you. Instead, I ask "What have you tried?" . Gerhard "So, What Have You Tried?" Paseman, 2016.10.29. $\endgroup$ – Gerhard Paseman Oct 29 '16 at 18:34
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    $\begingroup$ I guess the question is about two classes of integers: 1) all integer functions which satisfy $f(n! + m!)|f(n!)+f(m!)$ AND 2) all integer functions which satisfy $f(n + m)| f(n) +f(m)$ . $\endgroup$ – Pietro Majer Oct 29 '16 at 18:48
  • $\begingroup$ I think it's work with f(n)=n! $\endgroup$ – zeraoulia rafik Oct 29 '16 at 19:11

When $n=m$, $f(n)=kn$ then $f(m!)+f(n!)=k(m!+n!)=f(m!+n!)$

Therefore, $f(n)=kn$.

  • $\begingroup$ Why is $k$ constant? $\endgroup$ – Alex Degtyarev Oct 29 '16 at 23:09
  • $\begingroup$ It doesn't need constant. $\endgroup$ – Takahiro Waki Oct 29 '16 at 23:34
  • $\begingroup$ You are treating it as constant when you write $f(m!) + f(n!) = k(m! + n!)$. $\endgroup$ – LSpice Oct 29 '16 at 23:41
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    $\begingroup$ @TakahiroWaki, you must remember that you are in interactive website with professional mathematicians !!!!! $\endgroup$ – zeraoulia rafik Oct 30 '16 at 21:02
  • $\begingroup$ @L Spice I removed constant solution. see log. $\endgroup$ – Takahiro Waki Oct 31 '16 at 7:51

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