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Earlier, I asked a question; this is a question regarding an answer I received to it.

Apparently when $u^2 - 12p^2 = -3$, with $p$ a three digit prime, there is some sort of recursion of the values of $p$, causing $P(x+2) = 14P(x+1) - P(x)$, causing the first few values to be $1,13, 181,..$

Where did this recursion come from?

Note : $P(i)$ is referring to the $i$-th answer for $p$

J. W. Tanner
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justadumbguy
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  • After reading your first query, I spent a couple of minutes glancing at Eisenstein Integers. The topic seems complicated to me, which prompts me to wonder (re your previous query) how you could be confronted with such a question without any prior training. Was your original query from a contest or book/class? If a book/class, was any prior training involved before you were confronted with such a question? – user2661923 Dec 04 '20 at 03:20
  • it was from a book of questions, so no prior training was involved – justadumbguy Dec 04 '20 at 03:23
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    $u^2-12p^2=-3$ is an example of [tag:pell-type-equations] – J. W. Tanner Dec 04 '20 at 03:26
  • how does that imply the recursion? – justadumbguy Dec 04 '20 at 03:27
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    Notice that the comment of @J.W.Tanner indicated pell-type-equations (which I am totally unfamiliar with) as opposed to (simple) pell equations of the form $x^2 - Dy^2 = \pm 1.$ Personally, I think that you've walked into a minefield. For example, my number theory book asked a question that required me to study simple pell equations. I spent two weeks researching it before I felt like I had a reasonable intuition/knowledge of the subject. This is the risk inherent in attacking math questions without adequate prior training. – user2661923 Dec 04 '20 at 03:41
  • To clarify, not all $p$ in the series $1,13,181,2521, 35113,...$ are prime, and only one has $3$ digits – J. W. Tanner Dec 04 '20 at 04:22

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Let $u_1=3$ and $p_1=1$. Then $u_1^2-12p_1^2=-3$. I.e., $(u_1-2\sqrt3p_1)(u_1+2\sqrt3p_1)=-3$.

Now $7^2-12\cdot2^2=1$; i.e., $(7-2\sqrt3\cdot2)(7+2\sqrt3\cdot2)=1$.

Therefore, if $u_n^2-12p_n^2=-3$, then $(u_n^2-12p_n^2)(7^2-12\cdot2^2)=-3$.

I.e., $(u_n-2\sqrt3p_n)(7-2\sqrt3\cdot2)(u_n+2\sqrt3p_n)(7+2\sqrt3\cdot2)=-3$.

I.e., $([7u_n+24p_n]-2\sqrt3[7p_n+2u_n])([7u_n+24p_n]+2\sqrt3[7p_n+2u_n])=-3$.

I.e., $u_{n+1}^2-12p_{n+1}^2=-3$ with $u_{n+1}=7u_n+24p_n$ and $p_{n+1}=7p_n+2u_n$.

Furthermore, $p_{n+2}=7p_{n+1}+2u_{n+1}$ and $p_{n+1}=7p_n+2u_n$ and $u_{n+1}-7u_n=24p_n$ imply

$\dfrac{p_{n+2}-7p_{n+1}}2-7\dfrac{p_{n+1}-7p_n}2=24p_n$ or $p_{n+2}=14p_{n+1}-p_n$

(a recurrence involving $p$ and not $u$, which you seemed to want).

J. W. Tanner
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  • @justadumbguy J. W. Tanner has answered the specific question that you asked. Notice that the underlying question is being begged : how to determine the primitive solution of $u_1 = 3, p_1 = 1.$ Personally, I would have no clue how to determine the primitive solution from scratch. – user2661923 Dec 04 '20 at 03:48
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    my solution did not use Eisenstein integers, which involve $\sqrt{\color{red}-3}$ – J. W. Tanner Dec 04 '20 at 04:13
  • I agree, which I think is interesting. However, you must admit that identifying the primitive solution of (for example) $u^2 - 12p^2 = -3$ is different from attacking $x^2 - Dy^2 = \pm 1.$ The reason that I emphasize this is that my research into $x^2 - Dy^2 = \pm 1$ specifically emphasized that other pell-type-equations could be problematic to attack. – user2661923 Dec 04 '20 at 04:21
  • This relevant monograph is good – J. W. Tanner Dec 04 '20 at 04:23
  • Very interesting article, thanks. – user2661923 Dec 04 '20 at 04:37
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    @user2661923 If $u^2-12p^2=-3$ then $u$ is divisible by $3$, say $u=3v$, and $$(2p)^2-3v^2=1.$$ – Servaes Dec 04 '20 at 17:35
  • @Servaes Very nice. Hit my blind spot. – user2661923 Dec 04 '20 at 17:47