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Let $\left(\dfrac{a}{p}\right)$ denote the Legendre symbol. Prove that there are infinitely many primes $p$ such that $\left(\dfrac{p}{5} \right) = 1$.

Since there are infinitely many primes there must be infinitely many primes $p$ such that $\left(\dfrac{p}{5} \right) = -1,0,$ or $1$. How do we prove that $1$ is achieved infinitely many times?

user19405892
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2 Answers2

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Hint: Assume by contradiction that primes of the form $5k+1$ are finite, $p_1,\ldots,p_m$,
then consider $$ \Phi_5(p_1\cdot p_2\cdot\ldots\cdot p_m+1) \tag{1} $$ where $\Phi_5(x)=x^4+x^3+x^2+x+1$.

By Lagrange's theorem, every prime factor of $(1)$ has to be $\equiv 1\pmod{5}$, but $p_1,\ldots,p_m$ do not divide $(1)$.

Jack D'Aurizio
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The method that requires little extra machinery is to assume there is a finite list, let $P$ be the product of all of them, then take $$ N = 20 P^2 -1. $$ Well, $N$ is not divisible by $2,5$ or by any of the primes dividing $P.$ Quadratic residues show that $N$ is not divisible by any prime $q$ with $(q|5) = -1,$ since $5 \equiv 1 \pmod 4$ and $(q|5) = (5|q).$ However, either $N$ is itself prime or it is divisible by some prime, either way there is a new prime $p|N$ with $(p|5)=1.$

Will Jagy
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  • Why do you think of it as cheap? I find this is a pretty standard / student friendly approach to the problem. – Imago Jun 19 '16 at 19:58
  • @Imago Thank you. Sometimes I write low budget. On the whole, small machinery requirement. It is not evident here, but I typically have very quick answers for algebraic number theory of quadratic fields, where I use binary quadratic forms. Examples would include class number for a given discriminant, where they are writing ideals with the Minkowski bounds and getting lost, and I have an immediate list of distinct forms/classes. – Will Jagy Jun 19 '16 at 20:08
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    @Imago recommend book https://en.wikipedia.org/wiki/Mythago_Wood wonderful – Will Jagy Jun 19 '16 at 20:12