I want to solve the diophantine equation $y^2+y=x^3-1$.  I thought about writing it as $(y-\omega)(y-\overline{\omega})=x^3$, where $\omega=(-1+i\sqrt{3})/2$. If we prove that $y-\omega$ and $y-\overline{\omega}$ are relatively prime in $\mathbb{Z}[\omega]$, then it follows that there is $u\in\{\pm 1,\pm \omega,\pm\omega^2\}$ and $a,b\in\mathbb{Z}$ such that $$y-\omega=u(a+b\omega)^3=u(a^3+b^3-3ab^2+3ab(a-b)\omega).$$ I think we may be able to solve it by considering all the possible $a,b$ for a given $u$.  We two cases:   Case 1: $u=\pm 1$. This case is impossible since we would have $1=\pm 3ab(a-b)$ as an equality of integers. Case 2: $u=\pm \omega$. Here we have that $a^3+b^3-3a^2b=\pm 1$ and $y=\mp 3ab(a-b)$. I don't know what to do.   To sum up, I would like to know how to prove that $y-\omega$ and $y-\overline{\omega}$ are relatively prime in $\mathbb{Z}[\omega]$ and how to deal with case 2.
The only solutions that I know of till now are $(x,y) = (1,1) \space , (19,7)$. We can note that: $$x^2-x+1 = y^3 \implies (2x-1)^2 = 4y^3-3$$ Thus, if odd prime $p \mid y$, then $(2x-1)^2 \equiv -3 \pmod{p}$ and thus, $-3$ is a quadratic residue. This implies that $p \equiv 1 \pmod{6}$ .  How can we further proceed into this problem?  Note: As mentioned in one of the links in the comments, if we instead replace $x$ by $x+1$, we get $x^2+x+1 = y^3$. Thus, the solutions for this is $(x,y) = (0,1) \space, (18,7)$.
Is the product of two non-holomorphic function always non-holomorphic? Suppose $h(z)$ is a complex function. I have noticed that $h(z) = f(z)\cdot g(z)$, where $f(z)$ and $g(z)$ are non-holomorphic. Can $h(z)$ be holomorphic?   Can a similar statement be made, if $f(z)$ is holomorphic, but $g(z)$ is not? High-Swing Cascode Current Mirror I am currently dealing with current mirrors and came across the circuit shown below, a High-Swing Cascode Current Mirror. I read that this implementation, as the name suggests, has the advantage of a high voltage headroom, since the minimum potential at the drain terminal of M4 is: $$ V_{D4\_min}=V_{DS4}+V_{DS2}=(V_{GS4}-V_{TH})+(V_{GS2}-V_{TH}) $$ So the drain potential at M4 only needs to be twice the overdrive voltage, in case of same transistor dimensions, in order to hold M2 and M4 in saturation. But cannot this be said for many other cascode current mirrors, since that behaviour simply occurs by "stacking" together two transistors (M2 and M4)? Additionally, why is there a connection from the gate of M1 to the drain of M3? I have a hard time understanding whats going on in this circuit, help is greatly appreciated.    EDIT: A bias voltage source is setting the potential at the gate of M3/4. Sorry its not shown in the circuit.  EDIT 2: I just read that the mentioned bias voltage is implemented by a MOSFET in diode configuration, . For further discussion lets call it M5. If $U$ is unitary operator then spectrum $\sigma(U)$ is contained inside the unit circle In a Hilbert space, let $U$ be a continuous operator which it unitary. Prove $\sigma (U)\subseteq \Bbb{S}^1$.   It is important for me to know how I am doing, and I didn't come by a clear explanation so it would be appreciated to have your correcting and guiding.  Attempt: $U^{*}U=UU^{*}=I$ and therefore $U^{-1}=U$. I have $$ \| U x \|^2 = \langle U x, U x \rangle = \langle U^* U x, x \rangle = \langle x, x \rangle = \|x\|^2 $$ and the same goes for $U^*$ and therefore $\| U \| = \| U^{-1} \| = 1$.   Looking at $U-\lambda$ I can take $$ -\lambda U(U^{-1}-{1\over \lambda}I) =U-\lambda I. $$ Clearly, for $\lambda \ne 0$, $-\lambda U$ is defined and is a linear operator. $(U^{-1}-{1\over \lambda}I)$ is invertible if $|\lambda|\ge 1$ and therefore $\sigma (A^{-1})=\sigma (A)\subseteq \{|z|\le1\}$. But from the RHS, $U-\lambda I$ is invertible if $|\lambda|\le 1$ and therefore $\sigma(A)\subseteq \Bbb{S}^1$. I have noticed my claims tend to have holes in them many times which I cannot spot. If it is substantially true, how can I make sure it is completely formal? I am trying to find the equation of characteristic curves and solution of the partial differential equation $$x\frac{\partial u}{\partial y}-y\frac{\partial u}{\partial x}=u$$ $u(x,0)=\sin(\frac{\pi}{4}x)$ .   What I did: $$\frac{dx}{-y}=\frac{dy}{x}=\frac{du}{u}$$ with  From first two equalities $$\frac{dx}{-y}=\frac{dy}{x}$$ by solving we get $x^2+y^2=c_1$ for some constant $c_1$. Now I am stuck with the third and rest of the equalities to come up with a suitable manipulation to solve the problem. How can I do this? Any help would be great. Thanks. Let $\{s_n\}$ and $\{t_n\}$ be sequences.  I've noticed this inequality in a few analysis textbooks that I have come across, so I've started to think this can't be a typo:  $\limsup\limits_{n \rightarrow \infty} s_n + \liminf\limits_{n \rightarrow \infty} t_n \leq \limsup\limits_{n \rightarrow \infty} (s_n + t_n) \leq \limsup\limits_{n \rightarrow \infty} s_n + \limsup\limits_{n \rightarrow \infty} t_n$.  I understand the right two-thirds of the equation, and can prove it.  I'm stuck on the far-left third.  I don't understand how that is true. How does TeX actually work? It's not clear to me how to choose Spells Known when creating a new character at a higher level.  Let's take the Bard as a working example first:  Bard level 2:   Spells Known: 5 Slots per Spell level: 1st &times; 3   Bard level 3:   Spells Known: 6 Slots per Spell level: 1st &times; 4, 2nd &times; 2   I'm interpreting this to mean that, if an existing 2nd-level Bard gains a new level (level 3), he's gaining 1 (one) new spell known. He can then choose the new spell known to be either 1st or 2nd, correct?  What I don't understand then, is the case of creating a new character starting at a higher level; say, creating a new Bard at level 3.  In that case you're forced to choose 6 spells, which must be from the list of 1st- and 2nd-level spells, correct?  When making this 3rd-level Bard from scratch, though, can I select 2nd level spells for all 6 spells known?  As another example, say you're creating a very high level Bard from scratch, starting at level 15. The book says your Spells Known is 19, and your spell slots are 1st × 4, 2nd × 3, 3rd × 3, 4th × 3, 5th × 2, 6th × 1, 7th × 1, and 8th × 1.  So, for those 19 Spells Known, you can choose any spells from 1st through 8th spell level. Am I free then to select as many spells as I can at 8th spell level (5 in number), working backwards, and in the end ignoring the 1st, 2nd, 3rd level spells entirely?  As another example, the Ranger doesn't know any spells at 1st level, nor have any spell slots, but the book says “You know two 1st-level spells of your choice from the ranger spell list.” Does it mean at 2nd level and above?  Long story short: When choosing Spells Known for a spellcasting character created above 1st level, am I allowed to choose my Spells Known at whatever spell level I like, just so long as I have at least one spell slot of that level?  I'm working on a D&amp;D 5e character generator and need to clarify this. In summer, should I close curtains during the day? Is my sugar daddy scamming me How exactly do DLL files work? There seems to be an awful lot of them, but I don't know what they are or how they work.  So, what's the deal with them? When is unbalanced data really a problem in Machine Learning? How do I pass JavaScript variables to PHP? Invertibility of the Product of Two Matrices I have shown that if two $n\times n$ matrices $A$ and $B$ are invertible, then $AB$ is also invertible. I am having problems though with showing that $(AB)^{-1}=B^{-1} A^{-1}$. I thought of using their left-multiplication transformations but it seems that this is only associative and not commutative so I can reverse $L_A L_B$ A limit question (JEE $2014$) The following is a JEE (A national level entrance test) question:  Find the largest value of the non-negative integer ( a ) for which:  $$ \displaystyle \lim_{x \to 1} \left( \dfrac{-ax + \sin(x-1) + a} { x + \sin(x-1) -1 } \right)^{\dfrac{1-x}{1-\sqrt x} } = \dfrac 1 4 $$  On solving this, we get $ a = 0 $ or $ a = 2 $. So we take the answer as $ 2 $. But the official answer key says that the answer is $ 0 $. The reason given (not in the official key - they simple contain the answer. I saw the reason in an "unofficial" solution to the question paper) is that if $ a = 2 $, then the term $\displaystyle \dfrac{-ax + \sin(x-1) + a} { x + \sin(x-1) -1 } $ tends to a negative value ( $-0.5$ ). So $ a = 0$.  BUT:  By putting $a = 2$ in the limit, and inputting that to wolframalpha (), the answer turns out to be $0.25$. So, by putting $ a = 2$ also, we get the same value of the limit - $0.25$.  How to solve this dispute? Also, could you please check with Mathematica? Unfortunately, I don't have access to it!  (This is very important as the change in the answer key will affect the rankings in a dramatic way) An  is defined as $p = q\times d$ (for two point charges $\pm q$ separated by a distance $d$).    What is the physical meaning of this quantity?  Why does the direction of the electric dipole moment go from the negative charge to the positive charge? Another way of saying this, why electric dipole moment is a vector quantity? I am having trouble with the following problem:   $f: N\times N\rightarrow N$ and $f(i,j)=2^{i-1}(2j-1)$. Prove that $f$ is a bijection thus $N\times N$ and $N$ are numerically equivalent.   Work: I am told that Euclid's Lemma will be useful to proving injection and also that for any positive integer $n$, $n$ is a product of prime numbers.   For injection, I set $f(a,b)=f(x,y)\rightarrow 2^{a-1}(2b-1)=2^{x-1}(2y-1)$.  Then I distributed across to get $2^ab-2^{a-1}=2^xy-2^{x-1} \rightarrow 2^ab-2^xy=2^{a-1}-2^{x-1} \rightarrow 2(2^{a-1}b-2^{x-1}y)=2^{a-1}-2^{x-1}$.   From here, I am unsure how to apply Euclid's Lemma to arrive at the injection result.  I am even more unsure how to prove surjection. How do I mail merge from Google Spreadsheet to a Google Document? Looking for a single word for 'blind worship' Is there a single word for 'blind worship' as in worshiping an actor blindly notwithstanding the bad performance? How is dark energy consistent with conservation of mass and energy? I initially thought that dark energy must in some way violate conservation of mass and energy since the component of the energy density of space that comes from dark energy is constant, and space is expanding. Therefore, as space expands the energy in the universe that comes from dark energy would increase. I presumed the source of this energy was not coming from the conversion of other types of energy to dark energy, so it must violate conservation.  I decided to Google this and came upon this article:    It says that dark energy does NOT violate conservation and quotes Carroll, Press, and Turner (1992):  "…the patch does negative work on its surroundings, because it has negative pressure. Assuming the patch expands adiabatically [i.e. without loss or gain of heat], one may equate this negative work to the increase of mass/energy of the patch. One thereby recovers the correct equation of state for dark energy: P = – ρ c2. So the mathematics is consistent."  Is there a way to explain this in layman's terms? (The blog attempted to do this, but it was very unclear to me.) More specifically, can you explain where my initial train of thought described above fails when I erroneously concluded that dark energy violates conservation?  Thank you. Do I need a license to create pdf files? I hope this is the correct place where I could ask this question. My mother is an accountant with a degree in economics. She works as a freelancer and she needs some licenses for her job.  The biggest problem is adobe acrobat standard, which costs 400€, quite a lot. I want understand if she must buy it to create pdf files or she can use some free (even for commercial use) programs that she has because of her job (the chamber of commerce provide some advantages to accountants).  She is actually using PDFCreator, which as I can read is free for business usage (open source also!!):   Thanks for any suggestion