0

I was reading the book Putnam and Beyond. When I tried to calculate some examples, I came across one that I did not calculate, so I looked at the solution, but I did not understand it.

Here is the task:

The sequence $a_1,\,a_2,\,a_3,\,...$ of positive integers satisfies $\gcd(a_m,a_n)=\gcd(m,n)$ for $m\neq n$. Prove that $a_n=n$ for all $n$.

Here is the solution:

For any integer $m$, we have $\gcd(a_m,a_{2m})=\gcd(2m,m)=m$, and so $m$ divides $a_m$. Then, it follows that for any other integer $n$, $m$ divides $a_n$ if and only if it divides $\gcd(a_m,a_n)=\gcd(m,n)$. Thus $a_n$ has exactly the same divisors as $n$. Hence it must equal $n$, for all $n$.

More specifically, I did not understand the second and third sentences.

Thank you in advance.

emacs drives me nuts
  • 10,390
  • 2
  • 12
  • 31
Tomáš Macháček
  • 291
  • It seems that you are more interested in other proofs than the cited proof, so that makes your question a dupe. But if you want to understand the cited proof then see my answer. – Bill Dubuque Feb 27 '20 at 22:43

2 Answers2

3

Once you have that $m|a_m$, another way to argue is
Assume there is $a_n$ with $a_n \neq n$. Then $a_n=kn$ for some $k \gt 1$. We also know $kn|a_{kn}$, which would give $\gcd(a_n,a_{kn})=kn \neq n$ as required so we have a contradiction.

Ross Millikan
  • 374,822
0

The proof uses the gcd universal property $\ d\mid a,b\iff d\mid(a,b)\ $ as follows

$$\begin{align} m\mid a_n &\iff m\mid\ a_m,\ a_n\ \ \ \ \, {\rm by}\ \ m\mid a_m\\ &\iff m\mid (a_m,a_m)\ \ \ \rm by\ \ gcd\ universal\ property\\ &\iff m\mid (m,n)\ \ \ \ \ \ \ {\rm by\ hypothesis, and}\ \ m\neq n\\ &\iff m\mid \, \ m,n \ \ \ \ \ \ \ \ \,\rm by\ \ gcd\ universal\ property\\ &\iff m\mid n \end{align}$$

So $a_n$ and $n$ have the same set $S$ of divisors $m$, so the same $\rm\color{#c00}{greatest}$ divisor $\, a_n = \color{#c00}{\max} S = n$

Bill Dubuque
  • 272,048