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4:50 AM
@Balarka:
$f:\Bbb N\to\Bbb N$, with $f(mn)=f(m)f(n)$, $f(2)=2$, and $f$ strictly increasing
Now there is only one solution ($f(x)=x$) but do we need the strictly increasing property? Can't you just use FTA?
Or, rather, why can't you?
 
 
5 hours later…
9:40 AM
@SohamChowdhury Without that property you could define $f(2^p)= 2^p$ and $f(q) = 1$ if $q$ is odd and then extend multiplicatively. Now if $m = 2^k \cdot a$ and $n = 2^\ell \cdot b$ where $a$ and $b$ are odd, then $f(mn) = f(2^{k+\ell} ab) = 2^{k+ \ell} = f(m)f(n)$.
'Cuz $f(ab) = f(a)f(b) = 1$
I don't understand what you mean by FTA. Fundamental theorem of algebra? How?
With strictly increasing property this is obvious, of course
 

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