Welcome to our community

Be a part of something great, join today!

Problem Of The Week # 322 - Sep 06, 2018

Status
Not open for further replies.
  • Thread starter
  • Moderator
  • #1

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,925
  • Thread starter
  • Moderator
  • #2

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,925
Congratulations to castor28 for his correct solution, which can be read below:

Assume that there are only finitely many such primes, and call them $p_1,\ldots,p_n$. Note that, as $p_1=5$, this set is not empty.

Consider the number $N=\left(\prod{p_i}\right)^2+1$. Because $p_i\equiv1\pmod4$ for all $i$, $N\equiv2\pmod4$. As $N>2$, this shows that $N$ is not a power of $2$ and has at least one odd prime factor $q$.

We have $\left(\prod{p_i}\right)^2\equiv-1\pmod{q}$, which shows that $-1$ is a quadratic residue modulo $q$. By the laws of quadratic reciprocity, this implies that $q\equiv1\pmod4$. However, none of the $p_i$ can divide $N$, and $q$ is a prime of the form $4k+1$ different from the $p_i$; this contradicts the fact that the set $\{p_i\}$ contains all such primes.
 
Status
Not open for further replies.