Prove without quadratic residues or Dirichlet' theorem that there are infinitely many primes in the form $$5n+1, \ \ \ \ \ n\in \mathbb{N}$$
My try:
Obviously $n$ has to be even, or $5n+1$ is even. We can use $n':=\frac{n}{2}$, then $$5n+1=10n'+1$$ This should increase the odds of the number being prime. Since $5n+1$ the number of primes is about $\frac{li(n)}4$. For $10n'+1$, it should be $\frac{li(n')}2$. Does this help at all? or should I stay with $5n+1$?
Let we consider numbers of the form $a^5-1$. For any prime $p$ that divides $a^5-1$, we have:
$$ a^5 \equiv 1\pmod{p}\tag{1} $$ hence if $p>5$ and $a\not\equiv 1\pmod{p}$ we have that the order of $a$ in $\mathbb{F}_p^*$ is five, hence $5\mid (p-1)$ or: $$ p\equiv 1\pmod{5}. \tag{2}$$ In order to prove that an infinite number of primes of the form $5k+1$ exist, it is sufficient to show a sequence of numbers of the form $a^5-1$ with the property that neither $2,3$ or $5$ divide any term of the sequence, and any two terms of the sequence are coprime.
Or we can just apply Szigmondy's theorem to the sequence: $$ b_n = 12^{5^n}-1, \tag{3}$$ giving that $b_n$ has a prime factor that does not appear in $b_1,b_2,\ldots,b_{n-1}$. Such "new" prime factor is for sure $\equiv 1\pmod{5}$ by $(1)$ and $(2)$, hence there is an infinite number of primes of the form $5k+1$.
A slight variation of Jack's answer, falling closer in line with Euclid's proof.
If $1+x+x^2+x^3+x^4\equiv 0\pmod p$ with $p\neq 5$ prime, then $p\equiv 1\pmod 5$. This is because if $x\equiv 1\pmod p$ then $1+x+x^2+x^3+x^4\equiv 5\not\equiv 0\pmod p$, but $x^5\equiv 1\pmod p$ so $x$ is of order $5$ modulo $p$, and hence $p-1$ is divisible by $5$.
Now let $p_1,p_2,\dots, p_k$ be any set of primes.
Then let $m=5p_1p_2\cdots p_k$ and let $n=1+m+m^2+m^3+m^4$. Then any prime $p$ which is a divisor of $n$ is not one of $5,p_1,\cdots, p_k$. And we have $m^5-1\equiv 0\pmod p$ so $p\equiv 1\pmod 5$ since $m\not\equiv 1\pmod p$.
This works really for any number $k$, not just $k=5$.
Theorem: Given positive integer $k$, there exists infinitely many primes of the form $kn+1$.
Proof: Let $\Phi_k(x)$ be the cyclotomic polynomial for $k$. Then:
$$\Phi_k(x)\equiv 0\pmod{p}$$
only if $p\equiv 0,1\pmod k$. (That's something to prove, definitely.) Then the above proof works by taking taking the products of any finite set of primes and $k$ to get $K$ and find a new prime of this form by finding a prime factor of $\Phi_k(K)$.
So, for example, if $k=4$ then $\Phi_4(x)=x^2+1$ and $\Phi_4(4K)=16K^2+1$ is only divisible primes of the form $4n+1$.
Or if $k=6$, then $\Phi_6(x)=x^2-x+1$, and any prime factor of $\Phi_6(6n)=36n^2-6n+1$ is $\equiv 1\pmod{6}$ and not a divisor of $n$. So if $p\mid 36n^2-6n+1$ then $$p\mid 4(36n^2-6n+1)=(12n-1)^2 +3$$ so $-3$ is a square modulo $p$, which is only true when $p\equiv 1\pmod 3$ or $p=2,3$.
