1
$\begingroup$

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

Improve this question
$\endgroup$
7
  • 3
    $\begingroup$ 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, 2016 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, 2016 at 18:02
  • 2
    $\begingroup$ 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, 2016 at 18:34
  • 1
    $\begingroup$ 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, 2016 at 18:48
  • $\begingroup$ I think it's work with f(n)=n! $\endgroup$
    – zeraoulia rafik
    Oct 29, 2016 at 19:11

1 Answer 1

Reset to default
-3
$\begingroup$

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

Therefore, $f(n)=kn$.

Improve this answer
$\endgroup$
6
  • $\begingroup$ 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

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .