Here's a problem on one of the past pages that was incompletely settled, i arrived at a contradiction, and i don't know if someone can show otherwise; Assume that if $\phi_n=2^{2^n}+1$ then g.c.d($\phi_n,\phi_m$) = 1 whenever $n$ is not equal to $m$. Here's my approach. Let $k$=g.c.d($\phi_n,\phi_m$), then since $k|\phi_n$ and $k|\phi_m$, it implies, $k|\phi_n-\phi_m$, hence

$k|(2^{2^n}+1)-(2^{2^m}+1)$, for all $n>m$

$k|2^{2^m}(2^{2^n-2^m}-1)$, now $k$ does not divide $k|2^{2^m}$ because $\phi_n$ and $\phi_n$ are both odd, therefore, $k|(2^{2^n-2^m}-1$, and hence, $k=2^r-1$ for some integer $r$, where $r|2^{2^n-2^m}$

  • $\begingroup$ "and hence "$k=2^r-1$" is not shown. Also, how does it lead to contradiction? The start is good, and from $k$ divides $2^{2^n-2^m}-1$ you can indeed get $k=1$. $\endgroup$ – André Nicolas Feb 24 '14 at 20:50
  • $\begingroup$ Look math.stackexchange.com/questions/123524/… It may help $\endgroup$ – Josué Tonelli-Cueto Feb 24 '14 at 20:51
  • $\begingroup$ That was a good proof $\endgroup$ – Adokwu Ondoma Feb 24 '14 at 21:08
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    $\begingroup$ The answer linked to uses group-theoretic language, because the question did. So let's write a more number-theoretic version. Suppose $m\lt n$, and let $k$ divide both. Then $2^{2^m} \equiv -1\pmod{k}$. But $2^{2^n}$ is obtained from $2^{2^m}$ by repeated squaring, so $2^{2^n}\equiv 1\pmod{k}$. Since $2^{2^n}\equiv -1\pmod{k}$, we find that $k$ divides $2$. But $k=2$ is impossible. $\endgroup$ – André Nicolas Feb 24 '14 at 21:12

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