# Infinitely number of primes in the form 4n+1 proof

Question: Are there infinitely many primes of the form 4n+3 and 4n-1?

My attempt: Suppose the contrary that there exists finitely many primes of the form 4n+3, say k+1 of them: 3,$p_1,p_2,….,p_k$

Consider $N$ = 4$p_1p_2p_3…p_k$+3 $N$ cannot be a prime of this form. So suppose that $N$=$q_1…q_r$ where $q_i∈P$

Claim: At least one of the $q_i$’s is of the form 4n+3:

Proof for my claim: $N$ is odd => $q_1,…,q_r$ are odd => $q_i$ ≡ 1 (mod 4) or $q_i# ≡ 3 (mod 4) If all$q_1,…q_r$are of the form 4n+1, then (4n+1)(4m+1)=16nm+4n+4m+1 = 4() +1 Therefore,$N=q_1…q_r$= 4m+1. But$N=4p_1..p_k$+3 i.e.$N$≡3 (mod 4), N is congruent to 1 mod 4 which is a contradiction. Therefore, at least one of$q_i$≡ 3 (mod 4). Suppose$q_j$≡ 3 (mod 4) =>$q_j=p_i$for some 1$\leq$i$\leq$k or$q_j$=3 If$q_j=p_i≠3$then$q_j$|$N$= 4$p_1…p_k$+ 3 =>$q_j$=3 Contradiction! If$q_j$=3 (≠$p_i$, 1$\leq$i$\leq$k) then$q_j$|$N$= 4$p_1…p_k$+ 3 =>$q_j=p_t$for some 1$\leq$i$\leq$k Contradiction! In fact, there must be also infinitely many primes of the form 4n+1 (according to my search), but the above method does not work for its proof. I could not understand why it does not work. Could you please show me? Regards #### Solutions Collecting From Web of "Infinitely number of primes in the form 4n+1 proof" Suppose$n>1$is an integer. We define$N=(n!)^2 +1$. Suppose$p$is the smallest prime divisor of$N$. Since$N$is odd,$p$cannot be equal to$2$. It is clear that$p$is bigger than$n$(otherwise$p \mid 1$). If we show that$p$is of the form$4k+1$then we can repeat the procedure replacing$n$with$p$and we produce an infinite sequence of primes of the form$4k+1$. We know that$p$has the form$4k+1$or$4k+3$. Since$p\mid N$we have $$(n!)^2 \equiv -1 \ \ (p) \ .$$ Therefore $$(n!)^{p-1} \equiv (-1)^{ \frac{p-1}{2} } \ \ (p) \ .$$ Using Fermat’s little Theorem we get $$(-1)^{ \frac{p-1}{2} } \equiv 1 \ \ (p) \ .$$ If$p$was of the form$4k+3$then$\frac{p-1}{2} =2k+1$is odd and therefore we obtain$-1 \equiv 1 \ \ (p)$or$p \mid 2$which is a contradiction since$p$is odd. There’s indeed another elementary approach: For every even$n$, all prime divisors of$n^2+1$are$ \equiv 1 \mod 4$. This is because any$p\mid n^2+1$fulfills$n^2 \equiv -1 \mod p$and therefore$\left( \frac{-1}{p}\right) =1$, which is, since$p$must be odd, equivalent to$p \equiv 1 \mod 4$. Assume that there are only$k$primes$p_1,…,p_k$of the form$4m+1$. Then you can derive a contradiction from considering the prime factors of$(2p_1…p_k)^2+1$. (There’s also an elementary approach to show that there are infinitely many primes congruent to$1$modulo$n$for every$n$, but that one gets rather tedious. (See: Wikipedia as a reference.)) Suppose that there are only fitely many primes of the form$4k+1$, say$p_{1}, …, p_{n}$, and consider$N =(p_{1} …p_{n})^{2}+ 1$.$N>p_{i}$, for$1 ≤i≤n$, hence$N$cannot be prime. Any number of the form$a^{2}+1 $has, except possibly for the factor$2$, only prime factors of the form$4m+1$.Since division into$N$by each prime factor of the form$4k+1$leaves a remainder$1$,$N$cannot be composite, a contradiction. Hence, the number of primes of the form$4k +1\$ must be infinite.

Actually there is an even more elegant proof that shows that this holds true for all primes>2:

∀ primes,p ∈P & p>2,∃ k∈Q| k=(p-1)/4 or k= (p+1)/4

Proof:
By definition every 4th integer, k ≥ 4, must have 2^2 (2 squared) as a factor. This means that every other even integer > 2 must have 2^2 as a factor.

For all primes p>2, the prime itself is odd and therefore p-1 and p+1 are both even.

Since p-1 and p+1 represent consecutive even integers, one and only one of them must have 2^2 as factors and thus (p±1)/4 is an integer. ∎

This applies to all primes greater than 2.