$\gcd (127,m)=1 \Leftrightarrow 127 \not | 2^p-1$. If $127 | 2^p-1$ then $2^p \equiv 1 \pmod{127} \Leftrightarrow 18p \equiv 0 \pmod{126} \Leftrightarrow p \equiv 0 \pmod{7} \Leftrightarrow p =7$ - contradiction with $k>1$ ($7$ - because $2^7 \equiv 1 \pmod{128}$ - minimal such number). So $\gcd (127,m)=1$. Prove what $127|2^{2^p-2}-1$ and $2^p-1|2^{2^p-2}-1$ $127|2^{2^p-2}-1 \Leftrightarrow 2^p-2 \equiv 0 \pmod{18} \Leftrightarrow 2^{p-1} \equiv 2^{6k} \equiv 1 \pmod{3^2}$ - true. $2^{2^p-2}-1=(2^{p \cdot \frac{2^{p-1}-1}{p}}-1)(2^{2^{p-1}-1}+1)$ and $(\forall k)2^p-1|2^{pk}-1$ . P.S. Post is edited.

$\frac{2^{m-1}-1}{127m}$ is for $k=1,m=127$ not an integer by $LTE.$ So some mistake in question?

SCP wrote: $\frac{2^{m-1}-1}{127m}$ is for $k=1,m=127$ not an integer by $LTE.$ So some mistake in question? $k>1$

Posted also there: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=328276&hilit=Turkey+NMO+2007

Too easy for 2nd round? I'm gonna use cyclotomic polynomials to solve it. Before I get into the solution, let me explain a little bit what it is. The $n$th cyclotomic polynomial is a polynomial with an integer coefficient represented as; $\Phi_n(x)$ and defined as; \[ \Phi_n(X) = \prod_{\substack{\gcd(k,n) = 1 \\ 1 \le k \le n}} \left( X - \zeta^k \right). \]These polynomials have lot's of very usefull properties. I am going to give you just one of them to solve this problem. $Lemma:$ For any integer $n$, we have \[ X^n - 1 = \prod_{d \mid n} \Phi_d(X). \]In particular, if $p$ is a prime then \[ \Phi_p(X) = \frac{X^p-1}{X-1} = X^{p-1} + X^{p-2} + \dots + 1. \] Now, the problem wants us to show that: ${127m}\mid {2^{m-1}-1}$ $\implies$ $(2^7-1)(2^p-1)\mid {2^{m-1}-1}$ $\implies$ ${\Phi_7(2)}.{\Phi_p(2)}\mid \prod_{d \mid {m-1}} \Phi_d(2).$ But we already have $m-1 = 2^p -2 = 2(2^{p-1}-1) \equiv 0 \pmod p$ $\implies$ $p \mid {m-1}$ and $m-1 = 2^{6k+1} -2 = 2(2^{6k}-1) \equiv 0 \pmod 7$ $\implies$ $7 \mid {m-1}.$ Since $p=6k+1 \geq 13 > 7$ $\implies 7p \mid {m-1}.$ $\Box$

Trumba wrote: $\gcd (127,m)=1 \Leftrightarrow 127 \not | 2^p-1$. If $127 | 2^p-1$ then $2^p \equiv 1 \pmod{127} \Leftrightarrow 18p \equiv 0 \pmod{126} \Leftrightarrow p \equiv 0 \pmod{7} \Leftrightarrow p =7$ - contradiction with $k>1$ ($7$ - because $2^7 \equiv 1 \pmod{128}$ - minimal such number). So $\gcd (127,m)=1$. Prove what $127|2^{2^p-2}-1$ and $2^p-1|2^{2^p-2}-1$ $127|2^{2^p-2}-1 \Leftrightarrow 2^p-2 \equiv 0 \pmod{18} \Leftrightarrow 2^{p-1} \equiv 2^{6k} \equiv 1 \pmod{3^2}$ - true. $2^{2^p-2}-1=(2^{p \cdot \frac{2^{p-1}-1}{p}}-1)(2^{2^{p-1}-1}+1)$ and $(\forall k)2^p-1|2^{pk}-1$ . P.S. Post is edited. $$127\mid2^{2^p-2}-1 \iff 7\mid 2^p-2.$$