Post Number:#8 by SANZEED » Sat Nov 10, 2012 8:51 pm
For Problem 2
Let us denote the statement with $P(m,n)$. Now,
$P(1,1)\Rightarrow 2f(1)|2\Rightarrow f(1)=1$ since $f(m)\in \mathbb N$
Let $p$ be a prime.
$P(p-1,1)\Rightarrow f(p-1)+f(1)|p-1+1=p$. Since $f(p-1)1>1$ we must have $f(p-1)=p-1$.
Now, $f(p-1)+f(n)|p-1+n\Rightarrow (p-1+f(n))|(p-1+f(n))+(n-f(n))$.
So, $(p-1+f(n))|(n-f(n))$.
If we fix $n$ now then we can take arbitrarily large value of $p$, such that $p-1+f(n)>n-f(n)$. Still $(p-1+f(n))$ will divide $(n-f(n))$
so we must have $n-f(n)=0$ i.e. $f(n)=n\forall n\in \mathbb N$ which is indeed a solution.
Last edited by
SANZEED on Sat Nov 10, 2012 9:00 pm, edited 1 time in total.
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