Solving Diophantine equation without brute force
Prove $x^3 + 3 =4y(y+1)$ has no integer solutions
How fast does electricity flow? Why are there no animals in "The Matrix"? Compute the sum of the series $\sum_{n=1}^{\infty} \frac{1}{n \cdot 2^n}$ What would be the sum of the following series?  $$\sum_{n=1}^{\infty} \frac{1}{n \cdot 2^n}$$  Thanks C# datetime parse issue I can't read beta sites' chat logos! Here, you try:  If it weren't for the text shadow, it might as well be entirely invisible...  The cause is a CSS rule lurking in the transcripts for beta sites only:  #transcript-logo {   background: none repeat scroll 0 0 #273A5D; }   The problem is the #273A5D - a bluish colour not dissimilar to that used for the site name in the beta theme. Disabling that rule yields the much more manageable  I'll admit it is a little hard to read the "beta" in that shot. But if you had to choose which part was readable, wouldn't you choose the bit that tells you the name of the site?  Edit: It happens when actually in a chat room too! So I know I've heard this saying in American English before but I just can't quite find the original.  I have come up with several made up variations like:   I don't send the Navy (or whole army, or a battalion) when all you need is a diplomat. Don't buy the whole cow when all you need is milk.   What's the more commonly heard saying/phrase for this of expression? How can I make a program executable from everywhere What should I do if I want to be able to run a given program regardless of my current directory? Should I create a symbolic link to the program in the /bin folder? I have two partitions that I want to mount on startup: /dev/sda3   /mnt/devel  ext4    defaults    0   2 /dev/sda2   /mnt/excess ntfs    defaults    0   2  The ext4 partition mounts fine (owned by me, writable only by me), but the NTFS mounts owned by root with R\W permission for all. How to fix this? drwxr-xr-x  7 amanda amanda 4096 2012-03-14 19:07 devel drwxrwxrwx  1 root   root   4096 2012-03-14 22:38 excess I think I have managed to work this out, however there is no solution provided with the question so I thought it best to check.   My working:  Let $(x,y) \in (A \times B) \cup (C \times D)$  $\Rightarrow x \in (A$ or $C)$ and $y \in (B$ or $D)$  $\Rightarrow (x, y) \in (A \cup C) \times (B \cup D)$  $\Rightarrow (A \times B) \cup (C \times D) \subseteq (A \cup C) \times (B \cup D)$  Now let $(x,y) \in (A \cup C) \times (B \cup D)$  $\Rightarrow x \in (A$ or $C) $ and $y \in (B$ or $D)$  $\Rightarrow (x, y ) \in (A \times B) \cup (A \times D) \cup (C \times B) \cup (C \times D)$  $\Rightarrow (A \cup C) \times (B \cup D) \nsubseteq (A \times B) \cup (C \times D)$  $\Rightarrow (A \times B) \cup (C \times D) \not= (A \cup C) \times (B \cup D)$  Is this correct?  Thank you. I've been trying to find this darn series of books for ages, I read them back in early 2001-02 and can't really remember much of them. I think there were 10-ish books, it was an easy read but a long series. On early publications, the covers, were mainly yellow, with small pictures of whatever happens in that book.  The book began with two siblings, a brother and a sister, running in a forest. She conjures up what can only be remembered as a golem type creature. He then uses a high magic 'blast' and wipes out the golem, knackering himself. His sister laughs at him for being 'powerful' but expending himself too much on one big magic blast. The beginning I remember vividly, the rest, not so much.  The story continues as a conflict happens, I believe two worlds are involved? The brother and sister are split up, but this is where my memory becomes vague.  The mechanics of the world seems that there are gems, or stones, set in the backbone, which in turn give access to magic. When the person uses them, they can draw too much, which ends up exhausting them. The 'big bad evil king' kills their father, and destroys their farm early on in the story. Can infinitely many primes lie over a prime? For some reasons, homomorphism is a very hard area for me to make improvements. I've been hitting the brick wall for almost 2 hours.   Prove that no homomorphism exists from $Z_{16}\oplus Z_2$ onto $Z_4\oplus Z_4$.  Assume a homomorphism $\phi$ exists from $Z_{16}\oplus Z_2$ onto $Z_4\oplus Z_4$. Then $\phi$ is an isomorphism.  Note that the $\left | \ker \phi \right |=2$ The $\ker \phi$ is also a normal subgroup for $Z_{16}\oplus Z_{2}$. We want possible normal subgroup of order 2. I.e., 2 elements in each normal subgroups.  By Lagrange's theorem, the order of each element in a group divides the order of a group so the possible order of the elements are 1 or 2. If the elements has order 1, then the Ker \phi cannot have 2 elements.  Hence, $\ker\phi$ = $\left \{ (0,0),(8,0) \right \}$ ,$\left \{ (0,0),(0,8) \right \}$,$\left \{ (0,0),(8,1) \right \}$, possibly.  By the First isomorphism theorem:  $\phi: Z_{16}\oplus Z_2 \rightarrow Z_2\oplus Z_2$  $\Psi: G/\ker \phi \rightarrow \phi\left ( Z_16\oplus Z_2 \right )=Z_2\oplus Z_2$  $g\ker \phi  \mapsto \phi(g)=\Psi(g\ker \phi)$  Any help is appreciated.  Edit: I know that $(2,0)$ has order $8$ in $Z_{16}\oplus Z_{2}$ but any elements in $Z_{4}\oplus Z_{4}$ does not have order $8$ which would have solved the question at the outset. But I would like a different route. Is doing two PhDs a good path? Why I think tension should be twice the force in a tug of war I'm going to provide my argument for why I think the tension in a rope should be twice the force exerted on either side of it.  First, let's consider a different example. Say, there is a person named A and a block in space. A pushes on the block with a force of 100 N. Then, the block will also push A with a force of 100 N by Newton's third law. Now, consider the case where instead of the block, there is a person B who is also pushing on A with a force of 100 N while A is pushing on him. A will experience a force of 100 N because he pushed on B, AND another 100 N because he is being pushed by B. Hence he will experience a force of 200 N. Similarly, B also experiences 200 N of force.  Now, back to the original problem. There are two people A and B in space with a taut rope (no tension currently) in between them. If only A is pulling and B is not, then I agree that the tension is equal to the force A exerts. This situation (in my opinion) becomes analogous to the above if B is also pulling. So, say both of them pull from either side with a force of 100 N. Then the rope at the end of B will pull B with a force of 100 N (this pull is caused by A). By Newton's third law, the rope will experience a pull of 100 N. But B is also pulling his end of the rope with 100 N. Therefore, the tension should be 200 N. Similarly, the end of the rope at A must pull A with 100 N of force (because B is pulling from the other side) and hence experience a force of 100 N itself by Newton's third law plus another 100 N because A is pulling on the rope.  Apparently, the answer is not this (according to my searching on the web). So, could anyone tell me why this reasoning is wrong? Thanks.  EDIT : So apparently people don't agree with my first example, leave alone the second. This is to the downvoters and the upvoters of the highest-rated answer: You all agree that if only A pushes B with a force of 100 N, then A and B both will get pushed by a force of 100 N in opposite directions, right? Then, in the case where B is also pushing with a force of 100 N, it doesn't make sense that the answer would be exactly the same. It doesn't seem right that no matter what B does, B and A will always experience the same force as they would have if B hadn't applied any force.  EDIT 2 : I'm going to provide here a link to a question that I posted:  According to the answer and the comments there, the reason as to why my first example is incorrect is different to the one provided here. So maybe you should all read the answer and the comments provided by the person and reconsider what you think. Parse DateTime with time zone of form PST/CEST/UTC/etc Show that $\int_{0}^{1}\{P(x)\}^{2}\,dx = (n + 1)^{2}\left(\int_{0}^{1}P(x)\,dx\right)^{2}$ If $H$ is a cyclic subgroup of $G$ and $H$ is normal in $G$, then every subgoup of $H$ is normal in $G$. Exercise 11, page 45 from Hungerford's book Algebra.     If $H$ is a cyclic subgroup of $G$ and $H$ is normal in $G$, then every   subgroup of $H$ is normal in $G$.   I am trying to show that $a^{-1}Ka\subset K$, but I got stuck. What I am supposed to do now?  Thanks for your kindly help. Convert all text from uppercase to lowercase and vice versa? Determining true mean from noisy observations I have a large set of data points of the form (mean, stdev). I wish to reduce this to a single (better) mean, and a (hopefully) smaller standard deviation.  Clearly I could simply compute $\frac{\sum data_{mean}}{N}$, however this does not take in to account the fact that some of the data points are significantly more accurate than others.  To put it simply, I wish to preform a weighted average of these data points, but do not know what the weighting function should be in terms of the standard deviation. Calculate limit with L'Hospital's rule I want to calculate the limit $\displaystyle{\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}}$.   I have done the following:    It holds that $\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}=\frac{0}{0}$.   So, we can use L'Hospital's rule:  \begin{align*}\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}&amp;=\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x} \\ &amp;=\lim_{x\rightarrow 0}\frac{\left (x^2\cos \left (\frac{1}{x}\right )\right )'}{\left (\sin x\right )'} =\lim_{x\rightarrow 0}\frac{2x\cdot \cos \left (\frac{1}{x}\right )+x^2\cdot \left (-\sin \left (\frac{1}{x}\right )\right )\cdot \left (\frac{1}{x}\right )'}{\cos x} \\ &amp;=\lim_{x\rightarrow 0}\frac{2x\cdot \cos \left (\frac{1}{x}\right )-x^2\cdot \sin \left (\frac{1}{x}\right )\cdot \left (-\frac{1}{x^2}\right )}{\cos x} \\ &amp; =\lim_{x\rightarrow 0}\frac{2x\cdot \cos \left (\frac{1}{x}\right )+\sin \left (\frac{1}{x}\right )}{\cos x}=\lim_{x\rightarrow 0}\left (2x\cdot \cos \left (\frac{1}{x}\right )+\sin \left (\frac{1}{x}\right )\right ) \\ &amp; =\lim_{x\rightarrow 0}\left (2x\cdot \cos \left (\frac{1}{x}\right )\right )+\lim_{x\rightarrow 0}\left (\sin \left (\frac{1}{x}\right )\right )\end{align*}   We calculate the two limits separately   $\lim_{x\rightarrow 0}\left (2x\cdot \cos \left (\frac{1}{x}\right )\right )$ :    \begin{equation*}\left |\cos \left (\frac{1}{x}\right )\right |\leq 1 \Rightarrow -1\leq \cos \left (\frac{1}{x}\right )\leq 1  \Rightarrow -2x\leq 2x\cdot \cos \left (\frac{1}{x}\right )\leq  2x\end{equation*} we consider the limit $x\rightarrow 0$ and we get \begin{equation*}\lim_{x\rightarrow 0} \left (2x\cdot \cos \left (\frac{1}{x}\right ) \right )=0\end{equation*}    How can we calculate the limit $\lim_{x\rightarrow 0}\left (\sin \left (\frac{1}{x}\right )\right )$ ?