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I know there are sundry questions like this pdf — and this

(10.) Prove that any positive integer of the form $4k + 3$ must have a prime factor of the same form.

Because $4k + 3 = 2(2k + 1) + 1$, any number of the form $4k + 3$ must be odd.
It can't have any factors of the form $4k = 2(2k) $ or $4k + 2 = 2(2k + 1)$ which are even
— so they must have forms $4k + 1$ and $4k + 3$.
Suppose that they were all of the form $4k + 1$. Multiplying two such forms yields $(4k+1)(4m+1) = 4(4km+k+m) +1$, another $4k+1.$
Thus $\Pi$ (factors of the form $4k + 1$) must be another $4... + 1.$
Thus $\Pi$ (factors of the form $4k + 3$) must have a prime factor of the form $4k + 3\quad (♯)$.

I still don't understand Elementary Number Theory — Jones — p28 — Theorem 2.9.

Prove by contradiction. Suppose that there are only finitely many primes of this form $4k + 3$, say $p_1, ... , p_k$. Let $\color{red}{m = 4(p_1 ... p_k - 1) + 3}$. Since $m$ is odd, and the only even prime is $2,$ so each prime $p$ dividing m is odd.

(1 — Red) Where did this choice of $m$ hail from — feels uncanny?

By reason of $(♯)$ overhead, $m$ must be divisible by at least one prime of the form $4k + 3$ - name it $p_i$. Thence $p_i$ divides $(4p_l ... p_k -m = 1) \implies p_i = \pm 1$, a contradiction because $p_i$ is prime.

(3) How can I prefigure to consider $4p_1...p_k - m = 1$, in order to instigate a contradiction?

(4) Why does the method of proof here fail for proving infinitely many primes of the form $4k + 1$? I tried http://math.stackexchange.com/a/391103/85100.

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You confused between "divide" and "is divisible by" for (3) –  Gina Jan 5 at 7:22
    
Sidenore: If you think that is uncanny, you have not seen Ramanujan's work. –  chubakueno Jan 5 at 7:38
1  
You claim that "all nonzero numbers divide 1". The correct version is "all nonzero numbers is divisible by 1". $p|1$ is saying that $p$ divide $1$. Only $+1$ and $-1$ divide $1$ (hence the contradiction since $p$ is prime); even though all number (including zero), is divisible by $1$. Just to be clear: 3 divide 6; 6 is divisible by 3. –  Gina Jan 5 at 8:04
    
@Gina thanks a lot! i fixed it. –  Dwayne E. Pouiller Jan 5 at 8:10
    
(4) a number of the form $4n+3$ must necessarily have a prime factor of that form, where a number in the form $4n+1$ can have all its prime factors in the form $4n+3$. There are other constructions that might work though, consider $4(p_1p_2 \cdots p_n)^2 + 1.$ –  user45878 May 22 at 19:31

1 Answer 1

  1. The need for $-1$ is that $4(p_1 p_2 ....p_k)+3$ would divisible by $3$. So to prevent this you subtract $1$.

  2. $$m = 4 (p_1 p_2 \ldots p_k) -4 + 3 = 4 p_1 p_2 \ldots p_k -1$$ One could have defined $m$ like this and shown that $m \equiv 3 \mod 4$. It is just a matter of taste

  3. No prime can divide $1$. $1$ is divisible only by $\pm 1$ and neither are primes.

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Thanks, but can you please amplify your answers 1, 2? Are you able to answer my other questions? –  Dwayne E. Pouiller Apr 7 at 14:40

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