Infinitely many primes p that are not congruent to 1 mod 5 Argue that there are infinitely many primes p that are not congruent to 1 modulo 5.
I find this confusing.  Is this saying $p_n \not\equiv 1 \pmod{5}$?
To start off I tried some examples.
$3 \not\equiv 1 \pmod{5}$
$5 \not\equiv 1 \pmod{5}$
$7 \not\equiv 1 \pmod{5}$
$11 \equiv 1 \pmod{5}$
$13 \not\equiv 1 \pmod{5}$
$17 \not\equiv 1 \pmod{5}$...
If this is what the question is asking I've come to the conclusion that this is true.   Either way, I've got no clue how to write this as a proof.

You can follow the Euclid proof that there are an infinite number of primes.  Assume there are a finite number of primes not congruent to $1 \pmod 5$.  Multiply them all except $2$ together to get $N \equiv 0 \pmod 5$.  Consider the factors of $N+2$, which is odd and $\equiv 2 \pmod 5$.  It cannot be divisible by any prime on the list, as it has remainder $2$ when divided by them.  If it is prime, we have exhibited a prime $\not \equiv 1 \pmod 5$ that is not on the list.  If it is not prime, it must have a factor that is $\not \equiv 1 \pmod 5$ because the product of primes $\equiv 1 \pmod 5$ is still $\equiv 1 \pmod 5$
