{$a_n$} $\in N$ .if $(a_m,a_n)=(m,n)$ when $m \ne n$,prove that $a_n=n $ for all $n$

Question

The fact is that I am not getting the question clear if $a_m=5$ and $a_n=17$ and if I take $m=3$ and $n=7$ .This doesn't hold true.Can someone help me to understand what the question is asking me ?

score 0 · Answer 1 · answered May 15 '20 at 10:44

0

There is counterexample to your statement. Let us take $$a_n=\begin{cases} n^2,&& \text{if $n$ is prime}\\ n,&& \text{otherwise}\\ \end{cases}$$

If $m$ is prime then $(a_m,a_{m!})=(m^2,m!)=m=(m,m!)$. Otherwise $a_m=m$, and $a_{m!}=m!$, so the statement is evident.

answered May 15 '20 at 10:44

Pavel Kozlov

919

I assume that $m!=n$ in the title was meant as “not equal”, not as faculty. – Martin R May 15 '20 at 10:59
ok, in that case classical proof works, of course – Pavel Kozlov May 15 '20 at 11:06

{$a_n$} $\in N$ .if $(a_m,a_n)=(m,n)$ when $m \ne n$,prove that $a_n=n $ for all $n$

1 Answers1