# special case of Dirichlet’s theorem on primes in arithmetic progressions

The special case of Dirichlet’s theorem for primes in arithmetic progressions for primes congruent to $1$ modulo $q$ where $q$ itself is a prime can be established by the following elegant modification of Euclid’s proof (http://planetmath.org/ProofThatThereAreInfinitelyManyPrimes).

Let $f(n)=\frac{n^{q}-1}{n-1}=1+n+n^{2}+\cdots+n^{q-1}$. Let $n>1$ be an integer, and suppose $p\mid f(n)$. Then $n^{q}\equiv 1\pmod{p}$ which implies by Lagrange’s theorem that either $q\mid p-1$ or $n\equiv 1\pmod{p}$. In other words, every prime divisor of $f(n)$ is congruent to $1$ modulo $q$ unless $n$ is congruent to $1$ modulo that divisor.

Suppose there are only finitely many primes that are congruent to $1$ modulo $q$. Let $P$ be twice their product. Note that $P\equiv 2\pmod{q}$. Let $p$ be any prime divisor of $f(P)$. If $p\equiv 1\pmod{q}$, then $p\mid P$ which contradicts $f(P)\equiv 1\pmod{P}$. Therefore, by the above $P\equiv 1\pmod{p}$. Therefore $f(P)\equiv 1+P+P^{2}+\cdots+P^{q-1}\equiv 1+1+1+\cdots+1\equiv q\pmod{p}$. Since $q$ is prime, it follows that $p=q$. Then $P\equiv 1\pmod{p}$ implies $P\equiv 1\pmod{q}$. However, that is inconsistent with our deduction that $P\equiv 2\pmod{q}$ above. Therefore the original assumption that there are only finitely many primes congruent to $1$ modulo $q$ is false.

## References

• 1 Henryk Iwaniec and Emmanuel Kowalski. , volume 53 of AMS Colloquium Publications. AMS, 2004.
Title special case of Dirichlet’s theorem on primes in arithmetic progressions SpecialCaseOfDirichletsTheoremOnPrimesInArithmeticProgressions 2013-03-22 14:35:38 2013-03-22 14:35:38 bbukh (348) bbukh (348) 9 bbukh (348) Theorem msc 11N13