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Let $\,q\,$ be a prime of the form $\,4\, m_q+3$.

I ask if it is always possible to find two primes $\,p_1$ and $\,p_2$ of the form $\,4\, m_p+1$ such that $$q=\sqrt{p_1+\sqrt{p_2+q}}$$ E.g. $$3=\sqrt{5+\sqrt{13+3}}$$ $$7=\sqrt{37+\sqrt{137+7}}$$ $$11=\sqrt{101+\sqrt{389+11}}$$ $$19=\sqrt{337+\sqrt{557+19}}$$ $$23=\sqrt{521+\sqrt{41+23}}$$

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    $\begingroup$ Probabilistic argument suggests it should be true for large $q$. However, I am not sure it's even known that there always exists a prime $p_2$ such that $p_2+q$ is a square. $\endgroup$
    – Dmitry Krachun
    Feb 5, 2020 at 21:25
  • $\begingroup$ Out of curiosity, I wrote & ran a fairly basic C++ program to check all of the solutions for the primes $q$ up to $107$. In addition to the $5$ you list, it found many more. In prime:numSol form, the program found $3:1$, $7:4$, $11:12$, $19:9$, $23:35$, $31:25$, $43:35$, $47:86$, $59:73$, $67:89$, $71:130$, $79:73$, $83:254$, $103:140$ and $107:326$. The number of solutions fluctuates quite a bit, but they are generally on an upward trend. This is only a heuristic check, but it does indicate there could always be at least one solution for each prime $q$. $\endgroup$
    – John Omielan
    Feb 6, 2020 at 2:58
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    $\begingroup$ I found that such primes $p_1$ and $p_2$ exist for each positive integer $q \equiv 3 \mod 4$ up to $10^6$ (whether prime or not). $\endgroup$
    – Robert Israel
    Feb 6, 2020 at 5:08
  • $\begingroup$ I believe it is an open problem whether there is a prime of the form $n^2-q$ (cf. Dmitry Krachun's remark). $\endgroup$
    – GH from MO
    Feb 6, 2020 at 8:39

1 Answer 1

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It is very likely that there are finite many solutions. Reason: The prime solutions $(x,y)=(p_1,p_2)$ of a quadratic equation $$ax^2+bxy+cy^2+dx+ey+f=0$$ Is an open problem.

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  • $\begingroup$ What is the open problem? $\endgroup$
    – user6976
    Feb 6, 2020 at 2:23
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    $\begingroup$ Hello Rebel, Welcome to MO! I have flagged your answer and wanted to explain why I did. While there are some possible edits that could be made here for the sake of clarity (eg addressing Mark Sapir's comment). To me, this still reads as a comment not an answer to the question. You could certainly address such concerns by clearly stating how the posed problem reduces to an open conjecture. Alternatively, after gaining enough reputation on the site, you are allowed to comment on questions. $\endgroup$
    – Neil Hoffman
    Feb 6, 2020 at 3:52

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