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

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 '16 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 '16 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 '16 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 '16 at 18:48
• I think it's work with f(n)=n! – zeraoulia rafik Oct 29 '16 at 19:11

## 1 Answer

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

Therefore, $f(n)=kn$.

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