http://www.mathlinks.ro/Forum/viewtopic.php?t=55870 pwns it

Is it just me or does Kalva's solution (http://www.kalva.demon.co.uk/canada/casoln/csol698.html) not make sense...

chess64 wrote: Is it just me or does Kalva's solution (http://www.kalva.demon.co.uk/canada/casoln/csol698.html) not make sense... Seems to make sense to me (maybe I'm missing something though). My solution:

Nice solution.

Clearly function is multiplicative. Function is specifically defined if we know values for $ p^k$ where $ p$ is a prime, and $ k\in N$ We know that for every $ n$ $ f(2^n)=f(2)*f(2{}^n{}^-{}^1)=f(2)*f(2)*f(2{}^n{}^-{}^2)=...=f^n(2)=2^n$ Does that mean that for every $ n$ $ f(n)=n$?

NapoleonXIV wrote: Function is specifically defined if we know values for $ p^k$ where $ p$ is a prime We can do better; a completely multiplicative function (such as the one given) from the positive integers is defined by its values at the primes. But note that for an odd prime $ p$ we have $ f(p - 1) < f(p) < f(p + 1)$ and if we have shown (inductively) that $ f(q) = q$ for $ q$ a prime less than $ p$ then the fact that $ p - 1, p + 1$ are both composite allows us to conclude. This is essentially rjt's solution. Edit: Follow-up: Is the condition that $ f(2) = 2$ necessary? What if we only have $ f(2) \neq 0$?

For any $n$, $f(2^{n}) = 2^{n}$... we have that between two powers of two the numbers are in order so are the $f's$ because $ f(m)>f(n) $ if $ m>n $ a simple strong induction will do it. We can also use that the groups {${2^{n-1} +1, ... , 2^{n} - 1}$} and {$f({2^{n-1} +1), ... , f(2^{n} - 1})$} are between the same numbers and in the same order, this fact plus the fact that they are all integers may be a complete solution, really don't know.

We have the stronger problem Find all $f: \mathbb{N} \to \mathbb{N}$ such that 1. $f(2)=2$ 2. $f(mn)=f(m)f(n)$ if $\gcd(m,n)=1$ 3. $f$ is increasing

solution to the original problem, NOT THE STRONGER ONE f(2.1)=f(2).f(1) , so , 2=2f(1) , so f(1)=1 now we use strong form induction let P(n): f(n)=n. let P(1) ,....,P(k) be true { base cases have already been checked} f(k.k)=[f(k)]^2 , i.e. , f(k^2)=k^2 also f(k)=k as per the problem condition , f(k)<f(k+1)<.....f(k^2-1)<f(k^2) so , clearly , f(k+1)=k+1 , ......., f(k^2-1)=k^2-1 must hold now our induction is complete.

mathbuzz wrote: ... f(k.k)=[f(k)]^2 ... Unfortunately, property 2. is only valid when $\gcd(m,n)=1$ and so you cant use it for $m=n=k>1$

pco wrote: mathbuzz wrote: ... f(k.k)=[f(k)]^2 ... Unfortunately, property 2. is only valid when $\gcd(m,n)=1$ and so you cant use it for $m=n=k>1$ why ?? the problem says that f(mn)=f(m)f(n) for all m ,n. s o ,please elaborate. i gave the solution to the original problem , not the stronger one.

Sorry, I did not see your sentence "solution to the original problem, NOT THE STRONGER ONE", although it was in red characters and great letters.