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This problem was largely inspired by this problem here.

There were many counterexamples given to the problem, such as a multiplicative function that maps primes to a permutation thereof.

However, if $f(x)$ is a strictly increasing function, then what happens then?

It appears that if such $f(x)$ existed, then $f(1)=\gcd(f(1),f(y))$. This implies that $f(1)$ would have to divide $f(x)$ for any integer $x$. So one thing I tried doing was setting $g(x)=\frac{f(x)}{f(1)}$. However, I was not able to proceed further.

So how does one find all $f(n)$ such that $\gcd(f(x),f(y))=f(\gcd(x,y))$, where $f(x)$ is a strictly increasing function?

Note that all the counterexamples given were not increasing. Any help would be appreciated.

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Chad Shin
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    Could you specify here exactly what your question is, so that you don't have to refer us to a different question? – Wojowu Jan 29 '16 at 12:14
  • In either case, I believe $f(n)=\text{sgn}(n)(2^{|n|}-1)$ is an increasing sequence with asked property. – Wojowu Jan 29 '16 at 12:22
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    Or, sticking to natural numbers, $f(n)=F_n$, the $n^{th}$ Fibonacci number. – lulu Jan 29 '16 at 12:45
  • Why doesn't $f(x)=2x$ work? – s.harp Jan 29 '16 at 15:48
  • It appears that I lacked foresight in asking this problem. I did not completely think it through. For all $a,b$ where $a,b$ are coprime integers, $f(n)=a^n-b^n$ would be such a solution. In fact, I think they are called strong divisibility sequences. – Chad Shin Jan 29 '16 at 16:46

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