Exercise 1.25

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

If $a^n - 1$ is a prime, show that $a = 2$ and that $n$ is a
prime. Primes of the form $2^p - 1$ are called Mersenne primes. For
example, $2^3 - 1 = 7$ and $2^5 - 1 = 31$. It is not known if there are
infinitely many Mersenne primes.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow} \def\gcro{\mbox{[\hspace{-.15em}[}} \def\dcro{\mbox{]\hspace{-.15em}]}} ewcommand{\D}{\mathrm{d}} ewcommand{\Q}{\mathbb{Q}} ewcommand{\Z}{\mathbb{Z}} ewcommand{\N}{\mathbb{N}} ewcommand{\R}{\mathbb{R}} ewcommand{\C}{\mathbb{C}} ewcommand{\F}{\mathbb{F}} ewcommand{ e}{\,\mathrm{Re}\,} ewcommand{\ord}{\mathrm{ord}} ewcommand{ }{\mathrm{N}} ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}} \begin{proof} Suppose $n>1, a\geq 0$, and $a^n - 1$ is a prime. As $0^n -1 = -1,1^n-1 = 0$ are not primes, $a \geq 2$. Since $(a^n-1) = (a-1) (a^{n-1} + \cdots + a^{i} + \cdots + 1)$, $a-1$ is a factor of the prime $a^n-1$, so $a-1 = 1$ or $a-1 = a^n-1$. As $a\geq 2$, and $n >1$, $a = a^n$ is impossible, thus $a=2$. If $n\geq 2$ wasn't prime, then $n = uv, 1 < u 1)$ is a prime, then $a = 2$ and $n$ is a prime. \end{proof}

Exercise 1.25

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Exercise 1.25

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