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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

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6  
It doesn't work, because it doesn't work. A product of numbers 1 mod 4 can't be 3 mod 4, but a product of numbers 3 mod 4 can be 1 mod 4. There are other methods. –  Gerry Myerson Nov 26 '12 at 12:00
    
@GerryMyerson can you show me a way to prove it? –  Amadeus Bachmann Nov 26 '12 at 12:10
    
Use the Dirichlet's theorem which states that for a pair of numbers a, b satisfying gcd(a,b)=1, then the series {an+b} must contain infinitely many primes. Your problem is pointed out by Gerry Myerson. –  lee Nov 26 '12 at 12:12
    
A prime divisor of $m^2+1$ for an integer $m$ is either $2$ or equal to $1$ ($\bmod$ $4$). Then your argument can be adapted a bit to show that a finite number of such primes is insufficient to produce all numbers of this form. –  WimC Nov 26 '12 at 12:31
2  
@lee, Dirichlet is truly overkill for this problem. –  Gerry Myerson Nov 26 '12 at 21:58

2 Answers 2

up vote 8 down vote accepted

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.

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Could you explain me how you deduce the equality in the last display? I don't get how it follows from Fermat. How do you exclude the righ hand side to be -1? –  Koenraad van Duin Sep 15 at 21:52
    
It's from the $(n!)^{p - 1}$. –  zscoder Sep 22 at 13:28

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.))

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Didn't you get a warning that you were overriding my edit? –  joriki Nov 26 '12 at 12:30

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