1

There are an infinite number of primes of the form 4N-1 and a finite number of primes of the form 4N+1.

How would you go about disproving this statement ?

user129967
  • 195
  • There are in fact infinitely many primes of each type. The usual proof that there are infinitely many of the form $4k+1$ uses some facts about the congruence $x^2\equiv -1\pmod{p}$. – André Nicolas Mar 01 '14 at 02:59
  • The first is true and the second false, so the statement overall is false. Refer here to a proof that the second statement is false http://math.stackexchange.com/questions/244915/infinitely-number-of-primes-in-the-form-4n1-proof – MT_ Mar 01 '14 at 03:02
  • Can you use Euclids method of proof to disprove the statement? – user129967 Mar 01 '14 at 03:07
  • Yes, but one has to show something to the effect that $(2p_1p_2\cdots p_n)^2+1$ cannot be divisible by a prime of the form $4k+3$. This takes some work. – André Nicolas Mar 01 '14 at 03:30
  • Ok its fine don't worry but I'm still unsure of the proof as the proof provided in the answer is incorrect – user129967 Mar 01 '14 at 03:32

2 Answers2

1

Suppose $p_1, p_2, ... ,p_N$ are all primes in the form $4k +1$.

$M=(2p_1p_2 ... p_N)^2 + 1$. $M$ is not divisible by $2$ or any number in $p_1, p_2, ..., p_N$. Since none of the factors of $M$ are in the form $4k+1$, they must be in the form $4k + 3$.

If $p$ is a factor of $M$, then $(2p_1p_2 ... p_N)^2 + 1 = 0 \mod p$, or $(2p_1p_2 ... p_N)^2 = -1 \mod p$.

Quadratic residue tells us $x^2 = -1 \mod p \iff p = 1 \mod 4$, but this is a contradiction because based on our earlier statement $p$ must be in the form $4k + 3$. Therefore $p$ must be in the form $4k +1$ and not part of $p_1, p_2, ... ,p_N$, so it's larger than $p_N$. From this we see there are infinite primes in the form $4k + 1$.

qwr
  • 11,362
1

One can prove that there are infinitely many primes of the form $4k+1$ by using a variant of the "Euclid" argument. We will need the following useful lemma.

Lemma: Let $m$ be a positive integer. Then any prime divisor of $4m^2+1$ is of the form $4k+1$.

Proof: Suppose to the contrary that the prime $p$ of the form $4k+3$ divides $4m^2+1$. Note that $\frac{p-1}{2}$ is odd.

Since $p$ divides $4m^2+1$, we have $(2m)^2\equiv -1\pmod{p}$. Then $$(2m)^{p-1}=((2m)^2)^{(p-1)/2}\equiv (-1)^{(p-1)/2}\equiv -1\pmod{p}.$$ This is impossible, since by Fermat's Theorem $(2m)^{p-1}\equiv 1\pmod{p}$.

The rest is as usual. Suppose to the contrary that there are finitely many primes of the form $4k+1$, say $p_1,\dots,p_n$. Let $$N=4(p_1p_2\cdots p_n)^2+1.$$ Then $N$ has a prime divisor $p$, and by the Lemma $p$ cannot be of the form $4k+3$. Thus $p$ is of the form $4k+1$, and it is easy to see that $p$ cannot be any of $p_1,p_2,\dots,p_n$.

André Nicolas
  • 514,336