Show that $a^2 + b^2 = c^3 $ has infinitely many solutions
Variation of Pythagorean triplets: $x^2+y^2 = z^3$
Installing 64-bit geoprocessing of ArcGIS for Desktop? I am trying to install 64-bit processing to ArcGIS 10.2, but the ESRI .  One similar question on this exchange  that shows an option to include this, but my setup installation program (which I downloaded) does not show this installation window, but instead goes straight to selecting features to install as shown below.  I can't find any other documentation for running 64-bit geoprocessing on ArcGIS version 10.2.  Can anyone show me how to do this? or is it best to just try to install the service pack 1 version of it? Prove that $u\cdot v = \frac{1}{4}||u+v||^2 - \frac{1}{4}||u-v||^2 \forall u,v \in \mathbb{R^n}$ I am trying to prove the above statement but I'm not sure if my proof is correct. My proof is as follows,  Given $u\cdot v$, we know by the C-E Inequality that $|u \cdot v| \leq ||u|| \ ||v||$  Consider $\frac{1}{4}||u+v||^2 - \frac{1}{4}||u-v||^2$  We can manipulate this as follows $$\frac{1}{4}(||u+v||^2 - ||u-v||^2) = \frac{1}{4}(||u+v||- ||u-v||)(||u+v||+||u-v||)$$ Using the triangle inequality the above can be written as follows, $$...\leq \frac{1}{4}(||u|| +||v|| -||u||+||v||) (||u|| +||v|| +||u||-||v||) = \frac{1}{4} (4||u||||v||) = ||u||||v||$$  As such both $u\cdot v$ and $u\cdot v = \frac{1}{4}||u+v||^2 - \frac{1}{4}||u-v||^2 \forall u,v \in \mathbb{R^n}$ are $\leq ||u|| ||v||$  I'm not sure whether I've proven this correctly because I ended up proving that both sides of the equation are less than or equal to something entirely different (the magnitude of v multiplied by the magnitude of u). Is this acceptable? I don't think it is, since if both can be less than or equal to the magnitudes multiplied then they do not necessarily have to be equal to each other. If this is the case, can someone give me a hint on how to move forward with this proof? Why does Python's itertools.cycle need to create a copy of the iterable? React on global hotkey in a Java program on Windows/Linux/Mac? A Java6 application sits in the system tray. It needs to be activated using a hotkey (e.g. Super-G or Ctrl-Shift-L etc) and do something (e.g. showing an input box).  How do I do that on:   Windows (XP or Vista) OS/X Linux (Gnome or KDE) Objects with transparent texture not always show transparency in 3dview For some reason objects don't show their transparency all the time.  In this image the circles in red are suposed to be transparent like the ones in green.    When I select the object in the back, all objects in front of it become opaque.    When I select the two objects that are opaque, then they become transparent as they sould be.    The display is in GLSL mode  These are the material and texture properties      I'm not worried about the final render, just the editing view. I would like to see the actual transparency at editing time only, because the scene is to be imported in a third party game engine.  Blender is v2.68   is the blend file Optimization when Cost Function Slow to Evaluate Show reference label when it is undefined I'm aware there are been several similar questions, and some great answers with hints on how to prove this, for example  and .  But I've never really seen a detailed proof of the absolute convergence of the Riemann zeta function in the half-plane $\Re(s)&gt; 1$.  I hope there's interest on a detailed proof of this for reference. Here is my attempt. Please, fill in any detail that might be missing, or point out any mistake!  On $\Re(s)=\sigma&gt; 1$, we have  $$\sum_{n=1}^{\infty}\bigg|\frac{1}{n^s}\bigg|=\sum_{n=1}^{\infty}\frac{1}{n^\sigma}$$  and  $$\frac{1}{n^\sigma}\leq\frac{1}{n^{1+\epsilon}}$$  for any $\epsilon &gt;0$, so by the direct comparison test, if   $$\sum_{n=1}^{\infty}\frac{1}{n^{1+\epsilon}}$$  converges absolutely, then so do $\sum_{n=1}^{\infty}|1/n^s|$ and $\zeta(s)$.  By the integral test, the convergence of $\sum_{n=1}^{\infty}1/n^{1+\epsilon}$ is equivalente to the finitude of the integral  $$\int_1^\infty \frac{1}{n^{1+\epsilon}}=\bigg[-\frac{1}{\epsilon x^\epsilon}\bigg]=\frac{1}{\epsilon}$$  which holds for any $\epsilon &lt; \infty$.  Any correction or improvement is welcomed! Thanks in advance. For a Google account, you are allowed to sync Contacts and People details. What is the difference between these two? What's the difference between 18 A & 18 Ah on a Battery What's the difference between 18 A &amp; 18 Ah on a 12 V Battery While deploying code in vsCode from sandbox to production I get the below error      sf:INVALID_OPERATION: INVALID_OPERATION: testLevel of NoTestRun cannot   be used in production organizations   As per the error, I need to change the value of testLevel but I am not able to find where and how can I change it to RunLocalTests  Could someone please help me to change the value of testLevel in vsCode Turning off and on images in figures Is it possible to prevent the contents of a figure from being rendered? I have a large book document, and to speed up compiling while working on a draft, I would like to have some way of ignoring the content of the images, but ensuring that the figure size remains.  In several cases I am using a line like:  \includegraphics[width=\linewidth]{figure.pdf}     so I do not know the actual size of the image. My Notes has Title-Task2 and Content is "Talk to customers"  When I run following query:  SELECT Title,Content FROM ContentNote  Output is:  ('Title', 'Task2'), ('Content','/services/data/v42.0/sobjects/ContentNote/0692x000000WFDKAA4/Content')])]  how to get my actual content from this? Linux: coloring bash prompt will break carriage return What is the terminology for data aggregated via summed totals versus data aggregated via means? The two types of data differ in that if you decide to decrease the temporal (time) resolution of the first type of data you take the mean of lower the resolutions. With the second you take the sum over the lower resolutions. Here is a concrete example  Gas Used (kWh), Outside Air Temperature (C), Time Resolution (Minutes) 100, 20, 20 140, 22, 20 120, 21, 20   Here the hourly (60 Minute) resolution data is clearly  360, 21, 60   and is found by averaging the temperature and summing up the gas used.  What I really wish to know is what the names for each of these types of data are called. I understand that this question is not the best but it is hard to ask a question about something whose names you do not know. If you have a better way of articulating the question then please go ahead and edit it. How do I find this limit: $\lim_{n\to\infty} \left[ n-{n\over e}\left(1+{1\over n}\right)^n \right]$? I'm working on a practice exam for my masters quals and I am having difficulties with the following limit. According to wolfram alpha, it's value is 1/2. Does anyone know how to find this limit?   $$\lim_{n\to\infty} \left[ n-{n\over e}\left(1+{1\over n}\right)^n \right].$$ Will deleting my question also remove the reputation awarded to answerers? One of my questions got closed, as it was very similar to some other posted questions (though I don't feel the linked questions answered what I was trying to ask).  There's no point to the question staying online, as it doesn't answer anything, so I was going to delete it - but if I do that will it remove the reputation of the helpful answerers I want to upvote? I let him do it.   and     I allow him to do it.   Why does the latter require to?  What are the "rules" of using to with an infinitive? When is it necessary? How come Dropbox gets an icon in Finder sidebar? Double decimal formatting in Java