Given  $x^3-y^3=xy+61$. Find all pairs (x,y) of positive integers which satisfies the given equation.  I tried solving this by following method-  $(x-y)(x^2+y^2+xy)-xy=61$  $(x-y)[(x-y)^2+3xy]-xy=61$  $(x-y)^3+3xy(x-y)-xy=61$  I don't know if doing this was in any means fruitful or not.   Remember $x&gt;y$, only then a solution is possible.
I'm trying to solve the following equation:  $$x^3-y^3=xy+61$$  I got:  $$(x-y)(x^2+xy+y^2)=xy+61$$  But I can't go any further. I am looking for a solution in integers.  I need some hints to proceed, please!
How do I get a Seraph Crystal? Anthology for children; contains story about friendly Martian on postapocalyptic Earth I'm looking for the title of an anthology of children's science fiction stories from the 1950s. The best of the bunch involved a Martian who visited earth, and realized we'd blown ourselves up in a nuclear war and had fallen into savagery. The Martian is a vegetarian who subsists on biscuits made by a machine he brought with him, and communicates by using one of his tentacles to feed messages direction to/from the brain of the person he's talking to.  Inadvertently left behind by his expedition, he initially befriended some children, and eventually managed to rebuild civilization. Presently he dies, but when another expedition comes a generation later, they find the civilization he rebuilt.  there were other stories, but that was the best one.   If I can't find the name of the book it was in, I'd be happy to settle for the name of the author and the title of the story. Why does my image have space underneath? The  claims that likelihood and probability are distinct concepts.     In non-technical parlance, "likelihood" is usually a synonym for "probability," but in statistical usage there is a clear distinction in perspective: the number that is the probability of some observed outcomes given a set of parameter values is regarded as the likelihood of the set of parameter values given the observed outcomes.    Can someone give a more down-to-earth description of what this means?  In addition, some examples of how "probability" and "likelihood" disagree would be nice. Can players on a PC and an Xbox play together in any way? I.e. can Xbox users connect to a (modfied?) PC server, or vice versa? Get full URL (including querystring) from an apex REST endpoint Proof that $\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \geq a+b+c$ I've tried getting everything on the left side and transforming it into something squared so that I can prove it's bigger or equals to 0 but I've been unsuccessful. I'm having a bit of trouble proving this with the propositional calculus rules.  $$p \lor (p\implies q)$$ Would someone mind helping me and showing which rules they've used with an explanation! Thanks! Why is AES unbreakable? I have quite a bad habit of copy-and-pasting my packages from one LaTeX document to another. Is there any way we can detect unused packages (i.e. \usepackage commands) so that they can be removed from the LaTeX document? Sometimes when I am debugging code in Eclipse it happens that although I can see and inspect class member variables without any difficulty I am unable to inspect the values of variables declared locally within functions. As an aside, any parameters to the current function lose their 'real' names and instead one sees their values listed in the Variables window as arg0, arg1, arg2, etc but at least the values are visible.  This is occurring at present in relation to classes defined within the core JDK. I have verified that the installed and current JRE is a JDK.  Is anybody able to shed some light on this behaviour? Finding the dual basis for the space of polynomials of degree $\leq$ 2 Logical explanation why exponentiation operation is not commutative and associative Considering Peano axioms we'll define addition, multiplication and exponentiation operations.  We can then prove that addition and multiplication operations are commutative and associative. The proof for addition operation is pretty straightforward but for for multiplication it a bit more difficult. For example, to prove associativity, we have to prove distributivity first.  Moreover, for the definition itself it is not obvious that commutativity should take place, because the definition is not symmetric.  $a \cdot 0 := 0$  $a \cdot S(b) := a \cdot b + a$  Here $S(b)$ is the "next" function from Peano axioms  This definition is not symmetric with respect to $a$ and $b$, so commutivity is not obvious.  For exponentiation operation neither commutivity nor associativity take place. And I am very curious about the logical explanation why that happens  NOTE: The explanation I am looking for is somehow similar to a Abel–Ruffini theorem which explains WHY there is no an algebraic solution for a general polynomial equation How to understand a volatility based ETF like VXX The ETF VXX is based on the volatility index VIX. If volatility goes up, VIX goes up and VXX goes up. Similar thing for the downside. Now when I buy an ETF like SPY, I can conceptually understand that I am buying a small portion of each of the underlying companies. So if an underlying company stock makes a move, the ETF SPY makes a move in  the same direction.  But when I buy VXX, what exactly is the underlying? It is VIX. I am unable to grasp the concept of 'volatility' as the underlying. I can visualise only stocks and commodities as the underlying because in my understanding, the unerlying needs to be something physical (let us say stocks are physical commodities for this example).  Can someone explain the concept of 'volatility' as the underlying?  Is the word 'underlying' even appropriate in this case or is the word 'underlying' only to be used for derivative products? Taking the second derivative of a parametric curve I understand that for the parametric equations  $$\begin{align*}x&amp;=f(t)\\   y&amp;=g(t)\end{align*}$$  If $F(x)$ is the function with parameter removed then $\displaystyle F&#39;(x) = \frac{\text{d}y}{\text{d}t}\big/\frac{\text{d}x}{\text{d}t}$  But the procedure for taking the second derivative is just described as " replace $y$ with dy/dx " to get   $$\frac{\text{d}^2y}{\text{d}x^2}=\frac{\text{d}}{\text{d}x}\left(\frac{\text{d}y}{\text{d}x}\right)=\frac{\left[\frac{\text{d}}{\text{d}t}\left(\frac{\text{d}y}{\text{d}t}\right)\right]}{\left(\frac{\text{d}x}{\text{d}t}\right)}$$  I don't understand the justification for this step. Not at all.   But that's all my book says on the matter then it launches in to plugging things in to this formula, and it seems to work well enough, but I don't know why.  I often find answers about question on differentials are beyond my level, I'd really like to get this, it'd mean a lot to me if someone could break it down. Determining the default Method used in optimization and root-finding algorithms When you pick up items in Skyrim it tells you the value of each item. However when I go and sell the items the merchants are only paying 30% of this value, at best.  Discounting any perks or other effects, how high does Speechcraft have to be before merchants start paying full value for items? Splitting .tif image into several tiles? I have an image of size 1GB (.tif), with the width and height 94000x71680. I would like to chunk this image into 20000X20000 tiles so that I can process them.   How can I do this? I'm trying to continue the numbering of examples in enumerate environments in beamer. For example, I'd like one slide to feature   Foo Bar   and the next slide   Zip Yadda   My strategy so far has been to define a counter that saves the enumerate counter at the end of the first environment and can then be queried to set the enumerate counter at the start of the second enumerate environment. (This is essentially the proposal also found at )  That works wonderfully so long as I don't try to uncover elements of the slide sequentially. The problem, of course, is that as each of the pdf-pages that make up the slide is created, the counter is advanced. Here's a MWE.  \documentclass{beamer}  \setbeamercovered{highly dynamic}  \newcounter{saveenumi} \newcommand{\seti}{\setcounter{saveenumi}{\value{enumi}}} \newcommand{\conti}{\setcounter{enumi}{\value{saveenumi}}}   \begin{document}  \begin{frame}[&lt;+-&gt;]   \begin{enumerate}   \item foo   \item bar%     \seti   \end{enumerate} \end{frame}  \begin{frame}[&lt;+-&gt;]   \begin{enumerate}     \conti   \item zip   \item yadda%     \seti   \end{enumerate} \end{frame}  \end{document}   As the second slide is progressively uncovered (pages 3 and 4 of the resulting PDF), the counter at the beginning of the enumerate environment is advanced each time. So page 3 of the PDF has the right numbering, (3) and (4) respectively, while the fully revealed page has the wrong numbering. Pictures follow:   on one of the slides    on the following one Of all the people we know, in the books, who were sent to Azkaban, the majority are Death Eaters (Dolohov, Lucius, Crouch Jr, Bellatrix and so on) or suspected to be involved with them (Sirius, Stan, et cetera). They were imprisoned for using the Unforgivable Curses, as well as for being accomplices of a seriously dark wizard. What other crimes are Azkaban-worthy, in normal circumstances? I'm looking for anything that's been stated in the books, or on Pottermore (basically, anything 'canon'), or anything one can conclude from information in the books. I think smaller crimes are punished by imposing a fine, but what about bigger ones? Surely murder and torture (i.e. crimes against humanity) aren't the only ones punishable by imprisonment? That certainly isn't the case in contemporary Muggle society. My thoughts Off the top of my head, I can name these crimes:  Being an unregistered Animagus1 That was what Hermione threatened Skeeter with.  Trespass and robbery2 That's how Sturgis Podmore was imprisoned.   Notes  Several of the arrests in the Half-Blood Prince and the Deathly Hallows were made by an authoritarian (if that's the right word on this case) government, in the time of war. They do not count as normal.  I don't mean that there should be an answer to this question, I was wondering if it was ever addressed in the books. I certainly don't expect a list of all people sent to Azkaban, and the reasons for doing so!     1 &quot;Well, yes,&quot; said Hermione calmly, taking a sip of her drink. &quot;Otherwise, as you very well know, I will inform the authorities that you are an unregistered Animagus. Of course, the Prophet might give you rather a lot for an insider's account of life in Azkaban.&quot;  The Beetle at Bay, Harry Potter and the Order of the Phoenix     2 Sturgis Podmore, 38, of number two, ... charged with trespass and robbery ... sentenced to six months in Azkaban.  Percy and Padfoot, Harry Potter and the Order of the Phoenix