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Given a prime $p \equiv3 \mod 4$ ($p\geq7$) then the polynomial $n^2+n+\frac{p+1}{4}$ returns prime numbers for all $n\in\{0,1,\dots,\frac{p-7}{4} \}$. I have a sketch of a proof for this statement, but I haven't been able to fill in all the details. I am considering $K=\mathbb{Q}(\sqrt{-p})$, with $\mathcal{O}_K=\mathbb{Z}[ \frac{1+\sqrt{-p}}{2}]$. The proposed steps are as follows:

First, no element $z\in\mathcal O _K$ has a norm $l$, with $l\leq \frac{p-3}{4}$ being a prime number.

Second, all prime numbers $l\leq \frac{p-3}{4}$ are inert in $\mathcal O _K$.

Third, the elements $n+\frac{1+\sqrt{-p}}{2}\in\mathcal O _K$ (with $n\in\mathbb Z$) cannot be divided by prime numbers.

Fourth, if a prime element $\pi \in\mathcal O_K$ is a divisor of $n+\frac{1+\sqrt{-p}}{2}$ (with $n\in\mathbb Z$), then the norm of $\pi$ is a prime number greater or equal than $\frac{p+1}{4}$.

Fifth, if $n\in\{0,1,\dots,\frac{p-7}{4} \}$, then the norm of $n+\frac{1+\sqrt{-p}}{2}$ is a prime number.

Sixth, $n^2+n+\frac{p+1}{4}$ is a prime number for all $x\in\{0,1,\dots,\frac{p-7}{4} \}$.

I have written the details for parts 3 (supposing there is a prime number that divides it, and obtaining a contradiction) and 4 (using the second and third statements). Part 6 is trivial (since the norm of $n+\frac{1+\sqrt{-p}}{2}$ is $n^2+n+\frac{p+1}{4}$) once part 5 has been proved.

However, I'm stuck with parts 1,2 and 5. Regarding part 5, I have tried bounding the norm for the case $n=\frac{p-7}{4}$, but to no avail.

EDIT: as I said in the comments, I forgot to mention that we are considering the cases in which $\mathcal O _K$ is a PID.

sn2222
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    For $p=23\equiv3\pmod4$, $n^2+n+\dfrac{p+1}4$ is not prime for $n=0, 1, 2, 3, $ or $4$ – J. W. Tanner May 23 '21 at 17:37
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    This can't possibly be true. Just take $n = 1$. Your claim is than that $2 + \frac{p + 1}4$ is prime for all $p\equiv 3\pmod 4$, $p>7$. That is never true if $p \equiv -1\pmod 8$. Am I misunderstanding something? – Mathmo123 May 23 '21 at 17:38
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    See this answer for links to complete proofs. – Bill Dubuque May 23 '21 at 17:42
  • What is the source of the sketch? – Bill Dubuque May 23 '21 at 17:44
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    Also 1, 2 and 5 are all false. E.g. $3$ is inert in $\mathbb Q(\sqrt{-127})$, and the norm of $0 + \frac{1 + \sqrt{-p}}2$ is $\frac{1+p}4$ which certainly need not be prime. – Mathmo123 May 23 '21 at 17:49
  • The sketch is part of a proposed exercise that I was handed in class. I forgot to mention that we are considering the cases in which $\mathcal O_K$ is a PID. The exercise gives the polynomials with $p=7,11,19,43,67,163$ as examples. – sn2222 May 23 '21 at 18:09
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    @sn2222 The assumption that $\mathcal O_K$ is a PID completely changes everything. See comment posted by BillDubuque. Also those $p$ are the only examples. – Mathmo123 May 23 '21 at 18:20
  • The first sentence of your question is still wrong. Since there are no answers yet, you should fix the problem statement instead of putting an errata right at the bottom. (There is no need to signal "Edit" in the question if it does not affect the answers, since everyone can view the edit history.) – user21820 Oct 23 '21 at 08:23

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