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In my particular case he had kind of built the NS up in his head and it seemed like a challenge to get her into bed (it wasnt lol) and then once the reality of a 24 year old NS (half his age) had worn off it was too late and Wifey WAS LONG GONE! and so he was stuck with her. I noticed that my STBNEX is still using the NS to triangulate and try to get a reaction from me. I am too busy escaping to give two shits and so I totally ignore him.I noticed he was bored quite quickly (within 6 weeks) as he would still try talk to me. Mind you he was still saying “he’d come back in a heartbeat” if Id take him back.I know he’s bored because he went to the devalue stage so quickly. Beating her and talking about her like shes absolute shit - and to me…his ex of all people. She would die if she knew what he says about her to me! She’s living on borrowed time, he’s future faking her - keeping her around until he can find NS. Which at his age when your broke, no job, no money, no nothing is not so easy.I’ve been gone 18 months and 2 days and they have “been together” 18 months. It was his birthday the other day and instead of spending the evening with her he was too busy texting and thinking about me. Yeah he was “so angry” but the reality is its cos he wanted to hear from us for his birthday - he didn’t and then when he rang me I told him I wanted him to sign the divorce papers AND passport for our daughter!HE LOST HIS *hit over that!Moral of the story - they bore easily with some people and not so easily with others. He and I were married for 14 years and at the end I used to say “you need to let me go and go find yourself a new victim” to which his reply was “you are my favourite victim - I’ll never let you go”. And as f’d up as that sounds I think the sentiment is true. I was grade A supply and now he’s suffering cos I’m a hard act to follow. I actually feel sorry for the new victim - she has no clue, believes everything that comes out his lying mouth and will waste a good part of her life pandering to a man who hates her because in his mind it’s her fault I’ve gone and he lost everything. Meanwhile she’s got kids that she left by the wayside somewhere along the journey….that she’ll never see while she’s with him….too sad.Who cares if Narcs are bored? Really they are the worst forms of life you will ever encounter. Get free and stay narc free - that’s the best idea you’ll ever have about Narcs!

I fought for my house.It was scheduled to be auctioned in a foreclosure sale May 20, 2016.It wasn't really MY house, actually. It was the house that legally belonged to my soon to be ex-husband.My future ex-husband had decided to walk away from all of our mutual bills, including our mortgage. Unable to afford this mortgage on my income alone at the time as well as all the other bills, I decided to stop paying on the mortgage.So here's how it went down:My ex-husband decided to go silent for 6 months. He had no interest in discussing adult matters, such as the divison of property.One night, as I sorted through yet another box of his belongings, I decided I had enough of his silent treatment. The box I looked through contained pictures of his college trip to Jamaica with his friends and I thought it would really be shitty if I just started tossing these things out. I messaged him letting him know.Surprisingly, he decided to put on his adult pants for a minute and come discuss the future of the property.The house had equity in it. His mom, who was onced a licensed realtor, suggested he begin the process of a short sale.The house had equity…this meant if we both got our heads out of the shitter and worked together to sell the house, we could split that equity, catch up on our own bills and maybe have enough of a down payment on our own places.My lawyer said this would be the smartest thing to do.His lawyer, however, stated that she did not specialize in property law and, with the help of my future ex-husband’s parents, recommended that he get a restraining order against me.This was the state of affairs at the end of March 2016.We decided to save the house.I looked into foreclosure laws and legal assistance and anything to delay the auction. I found that New Hampshire does not give a fuck about bank foreclosure, probably because most people in this state tend to be wealthy in comparison to other states. The rate at which houses are lost to foreclosure here is not a regular problem.In New Hampshire, there is no judicial process.If you are behind on your mortgage and your lender has set the time frame for bank repossession and posted the auction in a local paper for 3 consecutive weeks, they can hold a sale outside your home on date stated in paper and then move to evict you.Every lawyer I spoke with had recommended bankruptcy. That would not work because he had been fired from his job recently and would not be able to show that he could make bankruptcy payments.I didn't know what to do, so I recommended filing a motion to Enjoin the Foreclosure. I didn't know what Enjoin meant, but I figured it might delay the auction. I would later find that it meant we were suing the mortgage company.He halfheartedly filled out the form at the courthouse. The form lacked in specificity.I began researching the hell out of case law and opinions in the State of New Hampshire. I also researched any previous problems and complaints with the mortgage company. I studied all the paperwork he signed and tried to find a loophole.I was coming up with nothing.Actually, I was finding that people who were Pro Se loved to file motions to Enjoin foreclosures, basically arguing that their loan was transferred to another mortgage company during the time of ownership and because of the transfer, they did not owe anyone money for their home. Huh? They wanted their home for free? Because they stopped paying on it? And now it was someone else's fault? Because the bank transferred the loan? Well, that was the dumbest thing I ever heard. And the more I read, the more I learned that judges thought so too..I directed him to begin communicating with the mortgage company and to find out how much we owed to catch up on the mortgage.He had money in his 401k which could possibly cover the amount past due. He was terminated, so it was possible to cash out on this money with tax consequences, of course.He reluctantly contacted the mortgage company. His prior experience with their customer service was awful. They would tell him something and then not follow through. But he called them to gather more information, specifically how much to pay off the amount in default.And they promised to send him the information.The funny thing about mortgage companies is that they show no signs of urgency when they are about to reclaim your property…especially if there is equity.They never sent him the information they promised.As it got closer to May 20th, we were informed that there was a QDRO hold on his 401k.…It's good to know about the imposition of a Qualified Domestic Relations Order, or QDRO (cue-DRO), an exception to the 10% penalty on distributions from a qualified plan (but not an IRA).In case of a divorce, it's good to know about Qualified Domestic Relations Orders.A QDRO is often put into place as part of a divorce settlement, especially when one spouse has a considerably larger retirement plan balance than the other. The court determines what amount (usually a percentage, although it could be a specific dollar amount) of a retirement plan's balance is to be presented to the non-owning spouse. Once that amount is determined and finalized by the court, a QDRO is drafted and provided to the non-owning spouse, which allows him or her to direct the retirement plan custodian to distribute the funds in the amount specified.QDRO Rules Use of 401(k) in DivorceThis meant that his money was stuck until the we agreed to who got what. And our next scheduled court date was cutting this close.So we canceled our divorce. His lawyer tried to change his mind. She really wanted to divorce me because I was such a pain in her ass… My lawyer shook his head and kind of laughed a little to himself. He was a cool guy.Meanwhile, our case was remanded to federal court which meant that it could take a bit before it was heard…pushing us closer to the deadline.A lawyer from the mortgage company reached out to my husband(no longer future ex). He demanded to know how much money he had. My husband told him 8k. He said that would not be enough to cover the costs. My husband demanded to know what those costs were. The lawyer promised to send him the information.We never received any information.Instead, we received a response from the attorney for the mortgage company in federal court stating all the reasons that they were not obligated to remove the house from foreclosure. These reasons included; lack of communication between my husband and the mortgage company for a plan(stating that if he had reached out, he would have known who had he talked to and when), lack of money my husband would be able to acquire even if there were a reinstatement plan(he stated that he had 8k and so the call from the attorney was just to find out more ways to fuck us and not to come to some type of agreement), and lastly, that the company did not have to help by providing a reinstatement plan as was stated in the loan agreement.That pissed me off… And my response was so damn good that it changed the judges mind…and the mortgage company’s reluctance to follow through and work with us… as well as the attorney's douchebaggery.The mortgage attorney even requested to have the lawyer who was now respresenting my husband contact him…um… that was no lawyer. That was me. A very pissed off me who had nothing but time on her hands to make him look like a fool for underestimating the power of a pissed off Sara.Oh and my response prompted an opinion to be published in New Hampshire law.. In his objection to the motion to dismiss, however, he provides additional factual background about his efforts to resolve the mortgage default with Pacific Union. He also states that the divorce proceeding has ended with a reconciliation between Angell and … and that he would like to pay the amount required to reinstate the mortgage but has not been able to do so because Pacific Union has not provided him with the required payment amount.Pacific Union has represented to the court that the foreclosure sale was cancelled “in order to allow exploration of loss mitigation efforts.” Pacific Union also represents that the foreclosure sale has not been rescheduled.DiscussionIn the complaint, Angell seeks to enjoin the foreclosure sale of his home. The foreclosure sale has been cancelled and has not been rescheduled. Therefore, at present, there is no foreclosure sale to enjoin. Pacific Union moves to dismiss the action on the ground that Angell has no right to reinstate or modify the mortgage and that Angell has not stated any other cause of action. In his objection to the motion, Angell agrees that he has not fulfilled the reinstatement requirement in the mortgage but argues that is because of Pacific Union’s failure to provide the reinstatement amount. He also agrees that he has no right to modify the mortgage but argues that he is entitled to relief based on a new theory of promissory estoppel. Given Angell’s pro se status and the change in circumstances and theories since the complaint was filed, it is appropriate to give him an opportunity to amend the complaint to allege facts to support a new cause of action. See Fed. R. Civ.P. 15(a). In addition, it appears that Pacific Union may be working with Angell to resolve the matter, which would be a better way to end the case. Therefore, the motion to dismiss is granted, but Angell will be given leave to file an amended complaint to allege a new claim or new claims. Pacific Union may respond as appropriate under the Local Rules and the Federal Rules of Civil Procedure.Civil No. 16-cv-167-JD Opinion No. 2016 DNH 104Updated Outcome:They had promised to provide us with an amount to bring the loan current and then ignored every attempt to reach out to them, either by putting us through to Voicemail (left a message and it was not returned), hanging up on us, or telling us that it was in the mail. They did not want to help.Each time we talked to someone from the mortgage company, we recorded the date, time, and the name of the person. We also took notes on the results of the call.In my objection, I wrote of all the runaround that they were giving us and the failure to do one simple thing, provide us with the amount to pay in order to reinstate.They started to work with after we filed the objection citing promissory estoppel, even before the judge had ruled that we could reopen a case against them for failure to follow through.They cancelled the foreclosure proceedings and referred us to loss mitigation who provided us with the paperwork for a loan modification.We didn't have to pay the whole past due amount. We only needed to make $400/month payments for 6 months. If we continued to make these payments on time, then we could ask them to review it and, upon review, they would determine if they would allow us to reinstate the loan.They did allow us to reinstate.We didn't have to pay for their attorney, as they had threatened us with earlier in the process.Our interest rate went up a bit… but there is still equity in our house.And we have never been late on a payment since…

My initial thought when I saw this question was “It isn’t any more.” This turned out to be not quite true, in the sense that Pell’s equation is important for certain areas of research in cryptography, although I do not know of any current implementations of it that are actually used in practice. So, let me explain a little about why Pell’s equation was important historically, how it transitioned into an algebraic number theory problem, and why this new context may yet lead to new cryptographic applications.The history of Pell’s equation begins with the discovery that [math]\sqrt{2}[/math] is irrational, often credited to Hippasus of Metapontum (although, as with many ancient discoveries, it is difficult to say with any particular certainty). This was incredibly significant to the Pythagoreans and the ancient Greeks in general, as so much of their mathematics was based around the notion of ratios. If something cannot be expressed as a ratio of integers, how do you work with it? On the other hand, given that this is literally just the length of the diagonal of a square, how can you not work with it?One solution is to come up with a procedure that can find approximations—that is, we would like some way of producing integers [math]X,Y[/math] such that [math]|X/Y [/math][math]-[/math][math] \sqrt{2}|[/math] is as small as we might like. It is conjectured that the ancient Babylonians already had such a thing, which today we refer to as the Babylonian method of computing square roots. However, the Babylonian method has one major weakness: the approximations that it produces have very large denominators. If you don’t have computers, and especially if you don’t have a decimal system (which none of these ancient cultures did), this means that if you want a very good approximation, you need to do a lot of very ugly computation.From this perspective, there is a better way via the Pell equation. The idea is thus. Let [math]d[/math] be an integer that is not a square. If [math]\sqrt{d}[/math] were a rational number, we would be able to find integers [math]X,Y[/math] such that [math]X^2 [/math][math]-[/math][math] dY^2 = 0[/math] (since we can rewrite this as [math]X/Y = \sqrt{d}[/math]). Since it is irrational, this is impossible, but we can still ask that this difference be as small as possible—in particular, we might ask whether we might be able to find integers [math]X,Y[/math] such that [math]X^2 [/math][math]-[/math][math] dY^2 = \pm 1[/math]. This is nothing more than the Pell equation. Solutions to the Pell equation have the nice property that [math]|X/Y [/math][math]-[/math][math] \sqrt{d}| \leq 1/Y^2[/math], so they give good approximations even with small denominators.Both Greek and Indian mathematicians studied special cases of this equation: for example, we know that the Baudhayana sutras give a very good approximation of [math]\sqrt{2}[/math] (correct to five decimal points) using the solution [math]X = 577, \ Y = 408[/math] to [math]X^2 [/math][math]-[/math][math] 2Y^2 = 1[/math], and Archimedes gave his own approximations for [math]\sqrt{2}[/math], although he did not explain how he produced them. The first general solution to the Pell equation was given by Bhāskara II in 1150, and was then rediscovered 500 years later by European mathematicians, including Fermat. The name “Pell” got attached to the equation entirely by accident—John Pell was an English mathematician who revised a translation of a book by Swedish mathematician John Rahn, in which Rahn described (among other things) English mathematician William Brouncker’s solution to the problem. Euler mistakenly thought that Pell was the one who had come up with the solution, referred to it in his writings as “Pell’s equation”, thus accidentally naming it after a person who had literally nothing to do with it. There have been some past efforts to rename the equation after someone more sensible, but none of them have caught on.In any case, by the time that Euler was studying Pell’s equation, its importance as a computational tool had diminished to essentially nothing. Finding the smallest solution to Pell’s equation for a given integer [math]d[/math] takes on the order of [math]\sqrt{d}[/math] operations (most of them spent writing out the solution, which will generally be very long)—producing new solutions from these can be done, but requires multiplying very large integers together. Unless you are especially interested in rational approximations (which isn’t so important once you have modern decimal notation to use), it is far, far easier to use something like the Babylonian method, or the more modern Goldschmidt algorithm, described in Andrew Bromage's answer here. Those algorithms also have the advantage that they can be applied to all real numbers, not just square-free integers.To Euler, Lagrange, and the mathematicians that followed them, Pell’s equation was primarily an exercise in pure mathematics and developing number theory. However, the full importance of it in those fields would not become apparent until the development of algebraic number fields.In the 18th century, number theorists started researching algebraic number fields. There are many equivalent ways to define what these are, depending on your degree of mathematical knowledge. Recall that we say that a number is algebraic if it is the root of a polynomial with rational coefficients—e.g. [math]2^{1/3}[/math] is algebraic since it is a root of [math]X^3 [/math][math]-[/math][math] 2[/math]. With this, we may say that an algebraic number field is a collection of [math]K[/math] which satisfies any of the following equivalent conditions.[math]K[/math] is a collection of algebraic numbers that is closed under addition, subtraction, multiplication, and division (except by [math]0[/math]), and there exists some finite collection of algebraic numbers [math]x_1, x_2, \ldots x_n[/math] such that every element in [math]K[/math] can be written down as [math]q_1 x_1 + q_2 x_2 + \ldots q_n x_n[/math] for some rational numbers [math]q_1, q_2, \ldots q_n[/math].[math]K[/math] is a field that contains the rational numbers [math]\mathbb{Q}[/math] as a sub-field, and [math]K[/math] is a finite-dimensional [math]\mathbb{Q}[/math]-vector space.There exists some finite collection of algebraic numbers [math]x_1, x_2, \ldots x_n[/math] such that [math]K[/math] is the smallest field that contains them and all of the rational numbers.There exists some algebraic number [math]x[/math] such that [math]K[/math] is the smallest field that contains it and all of the rational numbers. (Yes, this really is equivalent to (3), even though it really seems like it shouldn’t be.)The first algebraic number fields that were studied were the quadratic ones—these are the fields that you get by adjoining [math]\sqrt{d}[/math] to the rational numbers. That is, choose some integer [math]d[/math] that is not a square, and define [math]\mathbb{Q}(\sqrt{d})[/math] to be the collection of all elements of the form [math]q_1 + q_2 \sqrt{d}[/math], where [math]q_1,q_2[/math] are rational numbers. For example, [math]\mathbb{Q}(\sqrt{3})[/math] consists of elements like [math]3 + 4\sqrt{3}[/math], [math]1/3 [/math][math]-[/math][math] \sqrt{3}/5[/math], and the like. You can check that this collection satisfies the conditions of being an algebraic number field via any of the definitions (1)-(4)—it is pretty straightforward. (The hardest part is proving that this collection is closed under division, which can be done by rationalizing the denominator.)Another example are the Gaussian rationals [math]\mathbb{Q}(\sqrt{-1})[/math], which Gauss used to give elegant proofs of basic results in number theory, such as Fermat's theorem on sums of two squares. Of course, since Fermat’s theorem deals with integer solutions to [math]a^2 + b^2 = p[/math], Gauss couldn’t work with the Gaussian rationals themselves—instead, he took a subset of them, called the Gaussian integers, consisting of all elements of the form [math]a + b\sqrt{-1}[/math], where [math]a,b[/math] are integers. The relation with Fermat’s theorem then becomes clear, since [math](a + b\sqrt{-1})(a [/math][math]-[/math][math] b\sqrt{-1}) = a^2 + b^2[/math], and so the problem of looking for integer solutions to [math]a^2 + b^2 = p[/math] becomes a question of whether a given integer [math]p[/math] factors inside the Gaussian integers.This basic technique of taking an “integer-like” subset turns out to be applicable to any algebraic number field: for any number field [math]K[/math], there exists a ring of integers [math]\mathfrak{o}_K[/math]. Now, I would love to give a definition of what this is in general, but unfortunately it is difficult to explain in a way that is both understandable and feels motivated—one could easily write a whole other Quora post just about that. So, rather than doing that, we’ll just consider a special case: if [math]K = \mathbb{Q}(\sqrt{n})[/math], where [math]n[/math] is a square-free integer such that [math]n [/math][math]-[/math][math] 1[/math] is not divisible by [math]4[/math], then [math]\mathfrak{o}_K = \mathbb{Z}[\sqrt{n}][/math]—here [math]\mathbb{Z}[\sqrt{n}][/math] consists precisely of everything of the form [math]a + b\sqrt{n}[/math], where [math]a,b[/math] are integers. (The way to intuitively understand this notation is you are adding to [math]\mathbb{Z}[/math], the set of integers, the additional element [math]\sqrt{n}[/math].) You can check for yourself that this collection is closed under addition, subtraction, and multiplication. It is not, however, closed under division, which means that we can meaningfully talk about divisibility.In fact, an important question in algebraic number theory is how does factorization work in rings of integers of algebraic number fields. We just saw an example of how that might be used—specifically, factorization in the Gaussian integers gives you information about which primes are sums of two squares. I talked a little bit about how defining prime factorization works for things other than the integers in my answer to Are there any prime fractions?—one of the big takeaways from that answer was if you want to understand primes, you need to understand units. Units are any elements [math]x[/math] such that there is another element [math]y[/math] so that [math]xy = 1[/math]. Equivalently, they are the elements that you can divide by freely.What does any of this have to do with the Pell equation? A lot, actually! Let’s consider the equation [math]X^2 [/math][math]-[/math][math] 2Y^2 = \pm 1[/math] for a second. In the algebraic number field [math]\mathbb{Q}(\sqrt{2})[/math], the ring of integers is [math]\mathbb{Z}[\sqrt{2}][/math]. Therefore, if [math]X,Y[/math] are integers, then we have the factorization [math]X^2 [/math][math]-[/math][math] 2Y^2 = (X [/math][math]-[/math][math] \sqrt{2} Y)(X + \sqrt{2}Y) = \pm 1[/math]. This proves that [math]X + \sqrt{2}Y[/math] is a unit of this ring. In fact, you can check that it goes the other way, too: if [math]X + \sqrt{2}Y[/math] is a unit, then [math]X^2 [/math][math]-[/math][math] 2Y^2 = \pm 1[/math]. In other words, solving this Pell equation is the exact same thing as finding the units of [math]\mathbb{Z}[\sqrt{2}][/math].Rephrasing it as a problem of finding units is good on two fronts: first, it gives a new way that the equation can be used, and second, it gives more convenient ways of obtaining solutions. In particular, units can be multiplied—so, if we have one solution to the Pell equation, we can multiply by itself to get a new one. Thus, for instance, if we define[math]\displaystyle X_n + \sqrt{2}Y_n = \left(-1 + \sqrt{2})\right)^n, \tag*{}[/math]then [math]X_n^2 [/math][math]-[/math][math] 2Y_n^2 = \pm 1[/math] for all integers [math]n[/math]. Amazing!More generally, if [math]n[/math] is square-free and [math]n [/math][math]-[/math][math] 1[/math] is not divisible by [math]4[/math], then the units in the ring of integers [math]\mathbb{Z}[\sqrt{n}][/math] exactly correspond to solutions to the Pell equation [math]a^2 [/math][math]-[/math][math] nb^2 = \pm 1[/math] via the factorization [math](a + b\sqrt{n})(a [/math][math]-[/math][math] b\sqrt{n})[/math].When we talk about factorization in one of these rings of integers, what we really hope is that you can uniquely factor into prime elements. This is sometimes true—Gauss proved that [math]\mathbb{Z}[\sqrt{-1}][/math] has unique prime factorization, which gives you a neat proof of Fermat’s theorem on sums of squares. You can prove that [math]\mathbb{Z}[\sqrt{2}],\mathbb{Z}[\sqrt{3}],\mathbb{Z}[\sqrt{6}][/math] and many others have unique prime factorization. But many rings of integers don’t allow you to uniquely factor things. The classic example is [math]\mathbb{Z}[\sqrt{-5}][/math], which gives you the example that [math]6[/math][math] = 2 \cdot 3 = (1 + \sqrt{-5})(1 [/math][math]-[/math][math] \sqrt{-5})[/math], even though none of [math]2,3, 1 \pm \sqrt{-5}[/math] can be factored further, and none of them are related to each by multiplying by a unit.This leads to an important question: given a ring of integers, can we figure out whether it has unique prime factorization? The answer is “yes”, and it amounts to compute something that number theorists call the class number. Essentially, the class number [math]h_K[/math] of a ring of integers [math]\mathfrak{o}_K[/math] measures how far away it is from allowing unique prime factorization—if [math]h_K = 1[/math], then you have unique prime factorization, but the larger [math]h_K[/math] is, the more obstructions there are. A complete explanation of the class number would require delving quite deeply into algebraic number theory (it is the sort of thing that is typically taught in graduate courses in the subject), and so I will omit it. However, I will note that the most efficient way that we know to compute the class number involves the Dedekind zeta function, from which you get the famous class number formula. In general, this is a complicated thing, but we can state it reasonably simply if [math]n > 0[/math] is a square-free integer such that [math]n [/math][math]-[/math][math] 1[/math] is not divisible by [math]4[/math], and we are considering [math]\mathfrak{o}_K = \mathbb{Z}[\sqrt{n}][/math]—to wit,[math]\displaystyle h_K = -\frac{1}{2\log(\eta)} \sum_{r = 1}^{4n [/math][math]-[/math][math] 1} \left(\frac{4n}{r}\right) \sin\left(\frac{\pi r}{4n}\right), \tag*{}[/math]where [math]\left(\frac{4n}{r}\right)[/math] is the Kronecker symbol, and [math]\eta[/math] is the smallest unit of [math]\mathfrak{o}_K[/math] greater than [math]1[/math].Here’s the thing: even if this looks complicated, this is actually really easy to compute if you know what [math]\eta[/math] is. But figuring out what [math]\eta[/math] is equivalent to finding a minimal solution to [math]X^2 [/math][math]-[/math][math] nY^2 = \pm 1[/math]—in other words, if you want to understand when do rings of integers of quadratic fields have unique prime factorization, you should be looking at the Pell equation!And that is how the Pell equation earned such high importance in algebraic number theory.Now, all of that is great, but a priori it sounds hopelessly theoretical and perhaps utterly divorced from any sort of applications. This is not quite true. The broader theory of algebraic number theory certainly has its applications in areas like computer science, and Pell’s equation sometimes comes along for the ride.One of the areas that algebraic number theory is particularly prominent is cryptography. There are a number of reasons for this. Number theory is full to the brim of problems that have been studied for centuries, but are still considered very hard. It is well-suited for computation—as Donald Knuth wrote in 1974,virtually every theorem in elementary number theory arises in a natural, motivated way in connection with the problem of making computers do high-speed numerical calculations,and computational number theory continues to be a very robust field of study to this day. All of this together makes number theory a prime candidate for looking for examples of things like one-way functions and zero-knowledge proofs which are so important to cryptography.Do real quadratic fields show up in any of this? Well, sort of. There has been a string of papers starting from the 1980s exploring the possibility of using quadratic fields in the Diffie-Hellman key exchange—I described this cryptographic protocol in my answer to Why should I care about group theory? from a group theoretic perspective, noting that you could run the protocol with any group that had some nice properties. Well, in 1988 Buchmann and Williams proposed a key-exchange system based on imaginary quadratic fields by replacing the original group used by Diffie and Hellman with the class group of an imaginary quadratic field [math]\mathbb{Q}\left(\sqrt{-n}\right)[/math]. In 1991, Buchmann and Williams wrote another paper titled Quadratic fields and cryptography, in which they explored using real quadratic fields instead. At this point, they were no longer using something that was quite a group, and so it was the first version of the Diffie-Hellman key exchange that was not based on the difficulty of the discrete logarithm problem. They then co-authored with Scheidler to write A key-exchange protocol using real quadratic fields in 1994, which started fleshing out the algorithms needed to actually implement this cryptographic scheme.Now, you might be wondering why they were interested in pursuing a cryptosystem that acted like Diffie-Hellman but has somewhat different properties—after all, regular Diffie-Hellman seems to work just fine, so why fix something that isn’t broken? The issue is that having all of our cryptography based on the conjectured difficulty of a few fixed problems has the obvious weakness that if one of those problems turns out not to be difficult, then most of our cryptography is suddenly broken without a replacement. This possibility is not simply doomsaying, since we know that there are quantum algorithms that solve some of the problems used as a basis for cryptography in an efficient way. Thus, it behooves us to explore possible replacements.Unfortunately, the scheme with real quadratic fields has never taken off, quite simply because no one has yet managed to get it to run fast enough to be practical. A lot of the trouble actually boils down to the class number we mentioned before—this proposed scheme will only be secure if the class number is large. Unfortunately, the class number is observed to be [math]1[/math] about [math]76\%[/math] of the time!But people keep trying. In 2006, Jacobson, Scheidler, and Williams published a new paper on An Improved Real-Quadratic-Field-Based Key Exchange Procedure. Who knows? Perhaps one day it will in fact be successful.If you know where to look, you can find computations of solutions to the Pell equation (often phrased as searching for units in the ring of integers of a real quadratic field—but, remember, this is actually the same thing) in other proposed cryptographic schemes. For example, elliptic curve cryptography was at one point proposed as a replacement to RSA as a cryptosystem that would be resistant to attacks by quantum computers—unfortunately, it turns out the original scheme could be attacked by quantum algorithms every bit as much as the integer factorization used for RSA. There are various proposed solutions to this problem—one of them is to use isogeny classes of elliptic curves, and when you start digging into the literature of how this would actually work (see, for example, Abelian Varieties with Small Isogeny Class and Applications to Cryptography), you see that computations of units in the ring of integers of a real quadratic field do show up in there.Unfortunately, these proposed solutions suffer from the same problem as the Buchmann-Williams approach: nobody knows how to get them quite fast enough to be practical yet, although they are getting closer.To the best of my knowledge, as of right now, there are no notable users of cryptosystems that use computations of units of real quadratic fields. However, it is entirely possible that this will change in the next couple of decades.