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

Question

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

Can you give any context why you are interested in this question? In the present form the question looks suspiciously like a competition problem. — Jan-Christoph Schlage-Puchta, Oct 29, 2016 at 17:57
ok , really i 'd like to study and know more about multiplicative integer function and to study it's periodicity — ßłặck Pěặřł, Oct 29, 2016 at 18:02
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. — Gerhard Paseman, Oct 29, 2016 at 18:34
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 $f(n + m)| f(n) +f(m)$ . — Pietro Majer, Oct 29, 2016 at 18:48

Takahiro Waki · Accepted Answer · 2016-10-29 23:32:47Z

-3

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

Therefore, $f(n)=kn$.

answered Oct 29, 2016 at 22:40

Takahiro Waki

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$\begingroup$ Why is $k$ constant? $\endgroup$
– Alex Degtyarev
Oct 29, 2016 at 23:09
$\begingroup$ It doesn't need constant. $\endgroup$
– Takahiro Waki
Oct 29, 2016 at 23:34
$\begingroup$ You are treating it as constant when you write $f(m!) + f(n!) = k(m! + n!)$. $\endgroup$
– LSpice
Oct 29, 2016 at 23:41
1

$\begingroup$ @TakahiroWaki, you must remember that you are in interactive website with professional mathematicians !!!!! $\endgroup$
– zeraoulia rafik
Oct 30, 2016 at 21:02
$\begingroup$ @L Spice I removed constant solution. see log. $\endgroup$
– Takahiro Waki
Oct 31, 2016 at 7:51

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