infinitude of primes of special form Any number prime can only be written as $10k + x$ for $x=1,3,7,9$ since others cannot be primes.
Is there a way to prove that there are infinitely many primes of the form $10k+3$ or $10k+7$?
 A: This comes from Dirichlet's theorem, but there is an elementary way to
prove that the set
$$A=\{n\in\Bbb N:n\equiv\pm3\pmod{10}\}$$
contains infinitely many primes. Prove that each number of the form
$$10N+3$$
has a prime factor $p$ in the set $A$. Now take $N$ to be the product
of some primes in $A$ but excluding $3$. Then $p$ will be a further
prime ($\ne 3$) in $A$.
A: Yes. Let $F$ be a finite set of primes of the form $10k+3$ or $10k+7$. Let $P$ be the product of the elements of $F\setminus\{3\}$ (which is simply $F$ if $3\notin F$) and let $N=10P+3$. Note that $N$ is even and that $5\nmid N$. On the other hand, not all prime factors of $N$ are of the type $10k+1$ or $10k+9$, because the product of several numbers of this type is again of this type. Therefore, $N$ has a prime factor $p$ which is of one the forms $10k+3$ and $10k+7$. Furthermore, $p\neq3$, because$$3\mid N\iff3\mid10P\iff3\mid P,$$which is impossible, since $P$ is a product of prime numbers, none of which is $3$.
So, $N$ has a prime factor $p$ which is of the form $10k+3$ (but distinct from $3$) or of the form $10k+7$. But then $p\notin F$, because\begin{align}p\in F\iff&p\in F\setminus\{3\}\\\implies&p\mid P\\\implies&p\mid10P\\\implies&p\mid3\text{ (because $3=N-10P$)}\\\implies&p=3,\end{align}which is false.
