# there are an infinite number of primes $\equiv 1\mathrm{@}\text{symoperators}mod\text{tmspace}+.1667em\text{tmspace}+.1667emm$

This article proves a special case of Dirichlet’s theorem, namely that for any integer $m>1$, there are an infinite^{} number of primes $p\equiv 1\phantom{\rule{veryverythickmathspace}{0ex}}(modm)$.

Let $p$ be an odd prime not dividing $m$, let ${\mathrm{\Phi}}_{k}(x)$ be the ${k}^{\mathrm{th}}$ cyclotomic polynomial^{}, and note that

$$ |

If $a\in \mathbb{Z}$ with $p\mid {\mathrm{\Phi}}_{m}(a)$, then clearly $p\mid {a}^{m}-1$ and thus $\mathrm{gcd}(a,p)=1$. In fact, the order (http://planetmath.org/OrderGroup) of $amodp$ is precisely $m$, for if it were not, say ${a}^{d}\equiv 1\phantom{\rule{veryverythickmathspace}{0ex}}(modp)$ for $$, then $a$ would be a root $modp$ of ${\mathrm{\Phi}}_{d}(x)$ and thus ${x}^{m}-1$ would have multiple roots $modp$, which is a contradiction^{}. But then, by Fermat’s little theorem, we have ${a}^{p-1}\equiv 1\phantom{\rule{veryverythickmathspace}{0ex}}(modp)$, so since $m$ is the least integer with this property, we have $m\mid p-1$ so that $p\equiv 1\phantom{\rule{veryverythickmathspace}{0ex}}(modm)$.

We have thus shown that if $p\nmid m$ and $p\mid {\mathrm{\Phi}}_{m}(a)$, then $p\equiv 1\phantom{\rule{veryverythickmathspace}{0ex}}(modm)$. The result then follows from the following claim: if $f(x)\in \mathbb{Z}[x]$ is any polynomial^{} of degree at least one, then the factorizations of

$$f(1),f(2),f(3),\mathrm{\dots}$$ |

contain infinitely many primes. The proof is to Euclid’s proof of the infinitude of primes. Assume not, and let ${p}_{1},\mathrm{\dots},{p}_{k}$ be all of the primes. Since $f$ is nonconstant, choose $n$ with $f(n)=a\ne 0$. Then $f(n+a{p}_{1}\cdot {p}_{2}\mathrm{\cdots}{p}_{k}x)$ is clearly divisible by $a$, so $g(x)={a}^{-1}f(n+a{p}_{1}\cdot {p}_{2}\mathrm{\cdots}{p}_{k}x)\in \mathbb{Z}[x]$, and $g(m)\equiv 1\phantom{\rule{veryverythickmathspace}{0ex}}(mod{p}_{1}\mathrm{\cdots}{p}_{k})$ for each $m\in \mathbb{Z}$. $g$ is nonconstant, so choose $m$ such that $g(m)\ne 1$. Then $g(m)$ is clearly divisible by some prime other that the ${p}_{i}$ and thus $f(n+a{p}_{1}\cdot {p}_{2}\mathrm{\cdots}{p}_{k}x)$ is as well. Contradiction.

Thus the set ${\mathrm{\Phi}}_{m}(1),{\mathrm{\Phi}}_{m}(2),\mathrm{\dots}$ contains an infinite number of primes in their factorizations, only a finite number of which can divide $m$. The remainder must be primes $p\equiv 1\phantom{\rule{veryverythickmathspace}{0ex}}(modm)$.

Title | there are an infinite number of primes $\equiv 1\mathrm{@}\text{symoperators}mod\text{tmspace}+.1667em\text{tmspace}+.1667emm$ |
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Canonical name | ThereAreAnInfiniteNumberOfPrimesequiv1modM |

Date of creation | 2013-03-22 17:43:02 |

Last modified on | 2013-03-22 17:43:02 |

Owner | rm50 (10146) |

Last modified by | rm50 (10146) |

Numerical id | 7 |

Author | rm50 (10146) |

Entry type | Theorem |

Classification | msc 11N13 |

Related topic | SpecialCaseOfDirichletsTheoremOnPrimesInArithmeticProgressions |