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$จงพิสูจน์ว่า \frac{2^{n-1}+1}{n} ไม่เป็นจํานวนเต็มทุกจํานวนเต็มบวก n$

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We claim that the only positive integer $n$ such that $\dfrac{2^{n-1}+1}{n}$ is an integer is $n=1$. Suppose on the contrary that there exists a positive integer $n>1$ such that $n\mid 2^{n-1}+1$.

Note that $n$ must be odd. If $p$ is any prime integer that divides $n$, then
$$\left(2^{\frac{n-1}{2}}\right)^2=2^{n-1}\equiv -1\pmod{p}\,.$$
Thus, $-1$ is a quadratic residue modulo $p$. Consequently, $p\equiv 1\pmod{4}$. This shows that $n\equiv 1\pmod{4}$.

We shall now prove that every prime natural number $p$ that divides $n$ must satisfy $p\equiv 1\pmod{2^m}$ for every positive integer $m$. The cases $k=1$ and $k=2$ have been established. We suppose that the claim is true for $m=k$, where $k\geq 2$ is an integer. Then, $n\equiv 1\pmod{2^k}$, whence
$$\left(2^{\frac{n-1}{2^k}}\right)^{2^k}=2^{n-1}\equiv -1\pmod{p}$$
for any prime natural number $p$ that divides $n$. We already know that $p\equiv 1\pmod{2^k}$. We need to show that $p\equiv 1\pmod{2^{k+1}}$.

Let $g$ be a generator of the multiplicative group $(\mathbb{Z}/p\mathbb{Z})^\times$. Then, $-1\equiv g^s\pmod{p}$ for some integer $s$. Since $p\equiv 1\pmod{2^k}$ and $-1$ is a $(2^k)$-th power modulo $p$, we see that $2^k\mid s$. If $p\not\equiv1\pmod{2^{k+1}}$, then $\dfrac{p-1}{2^k}$ is an odd integer, and so
$$-1=(-1)^{\frac{p-1}{2^k}}\equiv (g^s)^{\frac{p-1}{2^k}}=g^{(p-1)\,\frac{s}{2^k}}\pmod{p}\,.$$
Since $\dfrac{s}{2^k}$ is an integer and $g^{p-1}\equiv 1\pmod{p}$ by Fermat's Little Theorem, we obtain
$$-1\equiv \big(g^{p-1}\big)^{\frac{s}{2^k}}\equiv 1\pmod{p}\,,$$
which is a blatant contradiction. Therefore, $p\equiv 1\pmod{2^{k+1}}$ must hold.

From the result above, we arrive at a contradiction. This is because, for any prime natural number $p$, $p-1$ can be divided by $2$ only finitely many times. Therefore, $n=1$ is the only positive integer such that $n$ divides $2^{n-1}+1$.