elementary-number-theorynumber theorypell-type-equations
What is known about primes in solutions to Pell-type equations?
In particular, consider the negative Pell equation $x^2 – 5 y^2 = -1$.
As far as I've been able to check
(in the first $4000$ solutions) the only positive-integer solution with $y$ prime is $x=38$, $y=17$,
but I don't see any obvious reason why this should be the case.
You period $n = 1$.
The first convergent is $\frac{73}{12}$ which means your fundamental solution can be $$x = 73, y = 12$$ and indeed $$73^2 - 37(12)^2 = 1$$
Assuming your continued fraction is right, the second convergent is
$$6+\frac{1}{12+\frac{1}{12}} = \frac{882}{145} $$ and indeed $$ 882^2-37(145)^2 = -1 $$ What you had done wrong is you had a wrong idea of what constituted the "first" convergent, using $12$ instead of $6\frac1{12}$. In point of fact, the zero-th convergent would be $6 = \frac61$ and indeed $$ 6^2 - 37(1)^2 = -1 $$ is the fundamental solution for the equation with $-1$ on the right.
If $p^2− \frac{251934169}{4} q^2 = 1 $, then $4p^2− 251934169q^2 = 4 $, so that $q$ has to be even.
Let $q = 2y$ so that $4p^2− 251934169\cdot 4 y^2 = 4 $ or $p^2− 251934169 y^2 = 1 $ and this becomes a standard Pell equation.
After searching for a Pell equation solver that could solve this, I found http://www.numbertheory.org/php/pell.php which said that
the least solution of $u^2-251934169·v^2=1$ is (u,v)=(225271244990700943315267956280411800555634015051308090277884868604104962338540162508662780103851110134288923470733214492576705304921942447496877316432148755394149624610709778342253829532034182955585281785605125774062067042871533474548723189550080879461940755054931210785029037747926428769992333668074349218880978527807663394127680486193384843986997533304388501754725074957250497216933172269065473852746178843181608306140365685996039286536330244666823292214037380277106713895640581283369187712083172695225344399463019401537455882605714942067846468852520297222251755337316170041845823661624202782666250115734237819683042082985171526282800382222714576727626094980629885366352000131303132135359463108635521565410488112287912844686192672738472001947756812093648970418119040213989933215258586913997217342108856777957037627456499897383934170286644285878283503578561992303795946148570976156177701503287373115452930781765236053893727748799204417831966968237024161827983124601075826155762117660452380906950456102876389134502447701968296040064628681054852876640098371536019198418255253941528944307814328148190184143000799951889089084659644347638577183884192073467082108506523506372732340274528141456740473415969963650656773018158163222322270988303396424521240862140458260591486474043045092368034899520901292279521856198411310822116084681636806884280270173841934612885750400068863907732865349185857915555235934810570449154233256465716941270067608222665579808387449666859364948244303214692248884665160509209379252943392494272634976297214741913847533965479527154125323880743892254267831724482576916275662958095933387932932551059660582766222292883165493697748307219389789209133143531345455092189866850439804333199651752360242139683699342368412986247802850384829304599110060419091883992716219173015297451039403047743238488985860729267543533352329687314970438461416693387772967943131761604987413334536162985092790422123282237028624645785633489664726218957148609919685769428122222084852809562569214413965720771408729185099918424343296677568166613090433613515483031252901694323178595995462642028964386862010093420726769418355184234720946181775332737659395967587225532076407623909565596790910765885818561865412162538848363553668612714198691288861437047601157353591221118724976585902490459382998907534939868158790879001671152433044317432949761648923963614896693699457510715718361695625294634696966284878410574160929772400728092044256419270960522476326245255572559146175279700400659950509793788476378848175390369825427301273114735692514379330175924474553668351745584388982458068437797359188706503857643164095213477929089334214332342050356909188421399238606889543471922745845239071010651328302511512759263707483733323614395083737283232899508265127479172494312201913477981379407799842709252906180434716507141311051322982240657546127971114278462713116439134410377103157523161671298844486230796509326848933853071382190119442436807702852665446333225245223437120117269735297360569365484765265678599514845945070544634609637487148190908846353641866101791106162516313305822511866796089071796216876389782118935966244678804350508265429071175447438279751936690584484769952697448915336468581146844571603113602643709943866971797062017021207642033921312513697587999212884164536775768402206018046772509817554058217211657510926455226154119243190033360301765607288639334708843614852895130144204537103834220400026878753494472540566640552609407477709815083333874222841700725622211428144359211679202120655592804708419101143418091589880834113404679889876903100324371120800767782414611156383670868080765233711399358523011346075275028463738338742036397612554950891163082740807747720798395812396497715250419270439996118401688429363409408587403868552632449,
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Best Answer
Let $\alpha = (2 + \sqrt{5})$ and $\beta = (2 - \sqrt{5})$. Note that $\alpha \beta = -1$. The general solution is given by
$$y_n = \frac{\alpha^n - \beta^n}{\alpha - \beta} = \frac{\alpha^n -\beta^n}{2 \sqrt{5}}$$
for $n$ odd. (This follows from general theory and I could explain it but I suspect that you know --- you can use induction, for example). This is related to the fact that $\alpha$ is a unit in the ring $\mathbf{Z}[\sqrt{5}]$. However, what is secretly going on is that there is the larger ring $\mathbf{Z}[\phi]$ where
$$\phi = \frac{1 + \sqrt{5}}{2},$$
and in fact $\phi$ is the fundamental unit, and we have $\alpha = \phi^3$ and $\beta = -\phi^{-3}$. So
$$2 y_n = \frac{\phi^{3n} - (-\phi)^{-3n}}{\sqrt{5}}$$
is actually divisible by
$$ \frac{\phi^{n} - (-\phi)^{-n}}{\sqrt{5}} = F_n,$$
where $F_n$ is the $n$th Fibonacci number. (This divisibility takes place in $\mathbf{Z}[\phi]$, but the ratio $2y_n/F_n$ is a rational number which is an algebraic integer and thus an actual integer.) Hence $y_n$ is divisible by $F_n$ if $F_n$ is odd and by $F_n/2$ if $F_n$ is even (and the ratio is also $> 1$ for $n > 1$). This shows that $y_n$ is not prime as soon as $F_n > 2$, so for $n > 3$. Hence $y_3 = 17$ is the only prime value.
For more general Pell-type equations I think one is generally out of luck unless there are forced divisibilities as in this case, and looks similar to the primality or otherwise of the sequence $2^n - 1$.
Added: I guess for those who want a more elementary solution, one can observe (and prove by induction) that the $y$ are given by
$$\frac{1}{2} F_{6n+3} = \frac{F_{2n+1} \cdot (5 F^2_{2n+1} - 3)}{2}$$
and the RHS is easily seen to be prime only for $n = 1$ whence $F_9/2 = 34/2 = 17$.