Just out of curiosity : has the equation $$ x^r+y^r=z^r,\qquad(x,y,z)\in\Bbb Z^3,\quad r\in\Bbb R, $$ been studied? Any non trivial result for $r\in\Bbb R\setminus\Bbb N$ ?
We now that Fermat's last theorem is true so there are not positive integer solutions to $$x^n+y^n=z^n$$ for $n\in\mathbb{N}$ and $n&gt;2$.  But what about if $n\in\mathbb{R}$ or $n\in\mathbb{R}^+$?
Exclusive differences Profiles vs permission sets? I understand, the differences between Profiles and Permission Sets. I just wanna ask are there some configs that make Profile or Permission Set unique.  Are there things that could only be done via permission set and not via Profiles and vice versa ? What constitutes an observation/measurement in QM? Commas that are sentence flow-stoppers ... Can I leave them out? Any help with this problem is appreciated. It comes up in the context of Lusin's theorem, where it was assumed to be true.     Suppose $E$ is a closed subset of $\mathbb{R}$ and $f:E\rightarrow \mathbb{R}$ is continuous. There exists a continuous function $g: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x)=g(x)$ for $x \in E$ and $g$ satisfies $$\sup_{x \in \mathbb{R}} |g(x)| \leq \sup_{x \in E} |f(x)|\;.$$ How to prove sum of powers property of roots of unity? Let $V$ be a subspace of $\mathbb R^3$ generated by the vectors $\{(1,1,1),(0,1,1)\}.$ There are $3$ vectors $u_1=(0,0,1),u_2=(1,1,0),u_3=(1,0,1)$.Then which of the following sets are connected?  $1.\left(\mathbb R^3\setminus V\right)\cup\{(0,0,0)\}$  $2.\left(\mathbb R^3\setminus V\right)\cup \{tu_1+(1-t)u_3:t\in [0,1]\}$  $3.\left(\mathbb R^3\setminus V\right)\cup\{tu_1+(1-t)u_2:t\in [0,1]\}.$  $4.\left(\mathbb R^3\setminus V\right)\cup \{(t,2t,2t):t\in \mathbb R\}$  Now $V$ is generated by two linearly independent vectors so is like $\mathbb R^2.$ So $\left(\mathbb R^3\setminus V\right)$ will be disconnected but giving the $(0,0,0)$ with them will make it connected.  No. $4$, I think will be connected also because $\{(t,2t,2t),t\in \mathbb R\}$ is in $V$ and contains the point $(0,0,0)$ Now, any neighbourhood of this point will have points from both $V$ and $\left(\mathbb R^3\setminus V\right)$ and so the union is connected.I'm not sure about no. $2$ and $3.$ There are many examples of this, and I'd like to give a few:   A person who puts a lot of effort to help the community and earns reputation points. But that reputation is the motivation behind helping the community, not because they really want to. A student approaches a professor, and helps them in their teaching and doing research in order to have a good LOR (letter of recommendation). To a SEO, the content of the website must be good (well, this one is a little tricky, but you get the spirit.)   I'm not saying that the reputation/LOR/high-ranked is not deserved. It is. I just want to focus more on the selfish reason(s) behind that.  Any kind of answer can be accepted: single word, phrase, proverb or idiom. How do I perform a redirect to a custom URL after login? I am trying to redirect users to a custom URL after they logged in, but my code doesn't work.  How do I redirect users after they logged in? Petersen graph is not a Cayley graph How can I show that the Petersen graph is not a Cayley graph?  I don't know very much about Cayley graphs, I know that they are vertex-transitive, but so is the Petersen graph. It probably has to do with the group structure of $\Gamma$ in $Cay(\Gamma,S)$ (which is a group of order $10$, i.e., it is either the cyclic group $C_{10}$ or the dihedral group $D_{10}$). Then I am guessing we go for some sort of contradiction(?). My group theory is a bit rusty, I would very much appreciate someone showing me how to do this one! Where to say thanks? Where is the right place to say "thanks" to all those that answered my question? In a comment to my original post, or in a comment to the latest answer or where? Was the chip real or fake? I'm talking about the movie  Spoilers ahead.  Walter (Daniel Radcliffe) gave the task or blackmailed the Four Horsemen to steal a chip from a lab which can access any digital device. So they did steal the chip. And when they rescue Dylan Rhodes (Mark Ruffalo) from drowning in the water, Jack Wilder said that the chip is fake.     Jack: Wait something is not right.      Atlas: What's wrong?      Jack: It's fake.   Then they assumed that Walter did not know that the chip is fake. So they came with a plan to bluff. They pretended that they had the real chip. After they got caught by Walter, they gave the fake chip to Walter. Walter checked the chip and declared it as real. The Four Horsemen were surprised.     Walter: I'm sorry to interrupt. But I have to say congratulations and   thank you. It's real.      Lula: What? It's real!!!!!      Atlas: What? Is it real?      Jake: What is he talking about?   What happened? I didn't get it. What did I miss? Was the chip real? Was Walter pretending it to be real? If so, then why? If not, who switched the real chip with the fake one? Who had the real chip before? My images are blurry! Why isn't WPF's SnapsToDevicePixels working? I'm using some Images in my WPF applcation.  XAML:  &lt;Image Name="ImageOrderedList"        Source="images/OrderedList.png"        ToolTip="Ordered List"        Margin="0,0,5,5"        Width="20"        Height="20"        SnapsToDevicePixels="True"        MouseUp="Image_MouseUp"        MouseEnter="Image_MouseEnter"        MouseLeave="Image_MouseLeave" /&gt;   But, they appear fuzzy.  Why doesn't that SnapsToDevicePixels="True" line prevent this problem? Specific system of differential equations I have the following system of equations:  \begin{eqnarray}\frac{dx}{dt} = x(1 - x^2 - y^2) \\ \frac{dy}{dt} = y(4 - x^2 - y^2) \end{eqnarray}  I want to prove that if a solutions starts (at time $t = 0$) in the region   $$ G = \{(x, y) \in \mathbb{R}^2 : x &gt; 0, y &gt; 0 \text{ and } 1 &lt; x^2 + y^2  &lt; 4\} $$  that it remains there for all $t \geq 0$. I have already proven that, from $G$, a solution cannot cross the $x$-axis where $0 &lt; x &lt; 2$ or the $y$-axis where $0 &lt; y &lt; 2$ (since those line segments are orbits), but the pieces of the circles with radius 1 and 2 (where $x &gt; 0, y &gt; 0$) are the lines I haven't done yet. In this question we can focus on just the circle with radius 1 since the proof for the other circle is completely analogous to it.  In my book, there is a similar problem where they show that if a solution crosses a certain line, that solution attains its maximum or minimum value for $x$ and $y$, and then they derive a contradiction. However, in this case, for $n \geq 1$ the $n$-th derivative of $x$ equals 0 on the circle with radius 1. (If the second derivative were negative, $x$ would attain its maximum on the circle, a contradiction since $\frac{dx}{dt} &lt; 0$ in $G$, so $x$ is decreasing.)  I don't see why a solution can't cross the circle at all. On the circle, $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} &gt; 0$, so after crossing the circle, the solution would go back up to $G$.  Is the circle itself an orbit?  EDIT I think I have found the answer. Using polar coordinates, I found $$ \frac{dr}{dt} = r - r^3 + 3r \sin^2(\theta) $$ Now, $$ \frac{dr}{dt} &gt; 0 \Leftrightarrow 0 &lt; r - r^3 + 3r \sin^2(\theta) \leq r - r^3 + 3r = 4r - r^3 $$ since $0 \leq \sin^2(\theta) \leq 1$. Note that $0 &lt; r \leq 1$ satisfies $4r - r^3 &gt; 0$, and thus $\frac{dr}{dt} &gt; 0$. Therefore, if a solution goes from $G$ to the edge of the unit circle, $r$ must, on the edge, either be decreasing or constant (i.e., its derivative is less than or equal to 0), contradicting the fact that the derivative is positive.  Can someone tell me if this is correct? How can I suppress the year from appearing in the bibliography for entries of type @inproceedings?  In my specific case the year is implied in the booktitle (e.g. FOCS '09) so it is redundant when I need to save space.  I found  and the "second solution" looks promising but I am not sure how to translate it to only suppress the year for one type of entry. I am guessing \ifentrytype might be helpful. I am using biblatex+biber. Which word to use when we ask for help?  Some conditions:     We know that the person asked is able to do it. We don't know if the person asked is able to do it. Trouble with proving $A$ is an integrally closed domain $\Rightarrow$ $A[t]$ is integrally closed domain Connecting multiple screens to the 2016 MacBook Pro with Touch Bar I'm thinking about buying the new MacBook Pro with 4 USB-C ports.  I want to attach two screens, mouse and keyboard and ethernet to the MacBook Pro while using the fewest number of USB-C ports possible.  So what I need is an adaptor with:   two DP or HDMI ports (or one DP port with daisy chaining support) two or more USB ports an ethernet port   Best case would be a hub to which all devices are connected with a single connection to the MacBook Pro, including power supply. Worst case if attaching the MacBook Pro to the devices would require me to use all four of USB-C ports.  Given the answers up to far, I can tell that:   Daisy chaining might or might not work with the MacBook Pro (since it requires DP1.2 which is supported by the graphics chipset, but it's uncertain if it's supported by the MacBook Pro itself) I can attach the monitors with two cables as a last resort, leaving me with two ports for the remaining peripherals.   What I still don't know:   Would daisy chaining work with a hub or does it require the first monitor to be directly attached to the MacBook Pro? Does some kind of hub exist supporting attachment of all listed devices with a single upstream connection to the MacBook Pro? How do I install the Nvidia drivers? I just ordered the Nvidia GTX card. I have a dilemma, though. Should I keep using the driver which is available in "additional drivers" in Ubuntu, or should I install the driver from the Nvidia site?  So which driver is the best for me? A linear operator commuting with all such operators is a scalar multiple of the identity. The question is from Axler's "Linear Algebra Done Right", which I'm using for self-study.  We are given a linear operator $T$ over a finite dimensional vector space $V$. We have to show that $T$ is a scalar multiple of the identity iff $\forall S \in {\cal L}(V), TS = ST$. Here, ${\cal L}(V)$ denotes the set of all linear operators over $V$.  One direction is easy to prove. If $T$ is a scalar multiple of the identity, then there exists a scalar $a$ such that $Tv = av$, $\forall v \in V$. Hence, given an arbitrary vector $w$, $$TS(w) = T(Sw) = a(Sw) = S(aw) = S(Tw) = ST(w)$$ where the third equality is possible because $S$ is a linear operator. Then, it follows that $TS = ST$, as required.  I am, however, at a loss as to how to tackle the other direction. I thought that a proof by contradiction,  ultimately constructing a linear operator $S$ for which $TS \neq ST$, might be the way to go, but haven't made much progress.   Thanks in advance! steam fatal error "steam needs to be online to update, but was set to offline movies." I tried re-installing  it but it simply keeps giving me this error in a little window. When I close it, an identical error pops up some 3 seconds later without any new provocation.