Prove: The integer p-1 is a quadratic residue of an odd prime p if and only if p congruent 1 ( mod4).  That’s right?!
I came across this problem and I believe Lagrange's theorem is the key to its solution.  The question is:     Let $p$ be an odd prime. Prove that there is some integer $x$ such that $x^2 \equiv −1 \pmod p$ if and only if $p \equiv 1 \pmod 4$.   I appreciate any help.  Thanks.
In my online brokerage account, I want to buy a particular stock and I see the following:  Bid:  13.20 x200           Ask:  13.27 x1,000   the current stock price is 13.22.   Can someone explain what the bid and ask prices mean relative to the current price?  If I buy 1000 shares, why would I pay more? What is the purpose of Microsoft.SharePoint.Runtime.dll in CSOM? I know we need to use Microsoft.SharePoint.Client and Microsoft.SharePoint.Client.Runtime dll's reference in .NET managed CSOM application development. But can someone help me understand why we have to use Microsoft.SharePoint.Client.Runtime and whats it's purpose, in detail. or point me to a reference please. Thank you in advance Why is this true: The only orthogonal projection that is also unitary from $\Bbb C^n$ to $\Bbb C^n$ is the identity Can anyone explain me please how to see this statement: the only orthogonal projection that is also unitary from $\Bbb C^n$ to $\Bbb C^n$ is the Identity.  how can I prove formally that? or how can I see that only the Identity satisfy it? Prove that the product of two continuous functions is continuous Showing a function on $[0,1]$ is the zero function Plotting recursive function using pgfplots \addplot Is it possible to plot a recursive function using pgfplots \addplot command?  I tried like this;  \documentclass{article} \usepackage{pgfplots}  \pgfmathdeclarefunction{fac}{1}{   \pgfmathparse{(#1&lt;=1)+                 (#1&gt;1)*#1*fac(#1-1)   } }  \begin{document} fac(5) is \pgfmathparse{fac(5)}\pgfmathresult  \end{document}   But the result seems to be an infinite recursion, as the entire function is being evaluated. How does damage dealt increase as you level up I've recently taken over as GM in my group.  We are all new to RPGs except for our original GM, so they decided to run an adventure path so it would be easier to change GMs if they had to stop playing.  Along this path, we had one CR 4 encounter that took us an hour, almost killed our bloodrager and knocked our brawler unconscious. When we leveled up, our GM told us to increase the damage we did by "one level," so my 1d6 became 2d6 and my friend's 2d6 became 4d6.  I saw no problem with this as a player, but I was looking over the adventure path and later encounters that also have a CR of 4 could be ended in one turn by our bloodrager. This would make encounters really boring, and my goal is to avoid boredom.  My question is: As you level up, how does the damage you do increase? Does it go up like my GM said so that at level 3 my rapier does 3d6 and the bloodrager's earth breaker does 6d6, or does it go up by some other method?  I have read the PHB and I don't think that what the GM was saying is correct, but being a first-time player I wasn't sure if had just missed something, so I figured I should ask elsewhere. How to understand Cauchy's proof of AM-GM inequality(the last step) How can I find $\gcd(a^m+1,a^n+1)$ with $a,m,n$ positive integers? How can I find $\gcd(a^m+1,a^n+1)$ with $a,m,n$ positive integers?    I have this idea:  Let $d=\gcd(m,n)$. Then there exist positive integers $x,y$ such that $mx-ny=d$ (WLOG). We shall find $G=\gcd(a^m+1,a^n+1)$.  If $m,n$ are odd, then $d$ is odd, therefore one and only one of $x,y$ is even. We have:  $$a^{ny}(a^d+1)=a^{mx}+a^{ny}=(a^{mx}-1)+(a^{ny}+1)=(a^{mx}+1)+(a^{ny}-1).$$  If $x$ is even and $y$ is odd, then $a^{m}+1\mid a^{mx}-1$ and $a^{n}+1\mid a^{ny}+1$, therefore $G\mid a^{ny}(a^d+1)$. If $x$ is odd and $y$ is even, then $a^{m}+1\mid a^{mx}+1$ and $a^{n}+1\mid a^{ny}-1$, thus $G\mid a^{ny}(a^d+1)$. However, since $\gcd(a^m+1,a^{ny})=\gcd(a^n+1,a^{ny})=1$, so $\gcd(G,a^{ny})=1$, hence $G\mid a^d+1$. We also have $a^d+1\mid a^{m}+1$ and $a^d+1\mid a^{n}+1$, so $a^d+1\mid G$. Thus $G=a^d+1$.  If $v_2(m)=v_2(n)=v_2(d)=k&gt;1$, then there exist some odd numbers $m_1,n_1,d_1$ such that $m=2^km_1,n=2^kn_1,d=2^kd_1$. We shall have $m_1x-n_1y=d_1$, so one and only one of $x,y$ is even, and we can use the same argument when $m,n$ are odd, so $G=a^d+1$.  However, if $v_2(m) \neq v_2(n)$, I cannot find any solutions for this. I think that $G \in \{1,2\}$, but I cannot prove or disprove it. How can I find $G=\gcd(a^m+1,a^n+1)$ if $v_2(m) \neq v_2(n)$ ? Moreover, is there anything from my arguments that can be improved ? When is 6 a square in Zp To find when $6$ is a square I divided $6$ in to $2$ and $3$ by The Legendre symbol.  $$\frac{6}{p} = \frac{2}{p} \times \frac{3}{p}$$  I know that we can multiply the cases of when $2$ and $$ are squares because then it is (1)(1) by Legendre.   $2$ is a square when $p \equiv 1 \pmod{8}$   $3$ is a square when $p \equiv 1 \pmod{12}$  so $6$ is a square when $p \equiv 1 \pmod {24}$ ($96$, but the $lcm(8,12)= 24$)   but how do I find the cases when $2$ and $3$ are both NOT squares? because then $6$ would be a square then as well, $(-1)(-1) = 1$ by Legendre. Where can I most efficiently buy all items for the "Shut up and take my money" achievement? I have just updated Windows 7 to Windows 10.  My touchpad scolling (right-side) is not working anymore.    I have no idea how to fix this in Windows 10.  I have hit the start button and typed "synaptic", but have found nothing useful.  I used to have an icon on the taskbar that showed where I was touching the touchpad.  I used to be able to right-click that icon to change settings, but Windows 10 has not brought that icon over.  Also, the asus website has no synaptic touchpad drivers available for Windows 10.  I have tried the enaltech drivers, but upon the installation atempt I am told that my hardware is not correct. Is it possible to control a value in all of my materials with one click? Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$? Using \ttfamily with \bfseries (or how to enable bold in fixed-width font) I'm using listings package, and I've added basicstyle=\ttfamily\footnotesize. However, when I do that, the keywords from the embedded source-code are not bold anymore.  I figured that \ttfamily together with \bfseries does not work.  And then I found a , and I tried adding \usepackage[bold-extras], and it didn't work either. What does that package do? Should I just ignore it and move on?  And then I finally found , which may be a duplicate to this one, but was a lot harder to find. I also found , which got me more confused.  So, what should I do in order to get a bold fixed-width font for use within listings?  Using pdflatex from texlive-2010. List vs ArrayList vs Dictionary vs Hashtable vs Stack vs Queue? Positive operator is bounded For a real Banach space $X$ let $A:X\rightarrow X^*$ be a positive operator in the sense that $(Ax)(x)\geq 0$ for all $x\in X$. Show that $A$ is bounded.   I don't know how to do that, maybe it's an application of the closed graph theorem? What are the differences between Factor Analysis and Principal Component Analysis? Differentiability of $g:=f(\sqrt{x^2+y^2})$ for a $C^1$ function with $f'(0)=0$ NBHM $2008$ Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a continuously differentiable function such that $f'(0)=0$.  Define for all $x,y\in \mathbb{R}$,  $$g(x,y)=f(\sqrt{x^2+y^2})$$ Pick out the true statements :   $g$ is differentiable on $\mathbb{R}^2$ $g$ is differentiable on $\mathbb{R}^2$ if and only if $f(0)=0$ $g$ is differentiable on $\mathbb{R}^2- \{ (0,0)\}$   I do not have much idea of differentiation of multivariable calculus.  I just tried out some examples to get some idea..  I have to take some function whose derivative at $0$ has to be zero.  for simplicity I would take $f(x)=x^2$ Then I would get :     $g(x,y)=x^2+y^2$ which is differentiable on $\mathbb{R}^2$   This would suggest me that second option has more chances as $f(0)=0$ in this case.  I would now take $f(x)=x^2+1$ then :     $g(x,y)=x^2+y^2+1$ which is differentiable on $\mathbb{R}^2$ though $f(0)\neq 0$   So, I would now eliminate second option and third option also...  So, It is now got confirmed that i have to work out to "Prove" that first option is correct.  Now, I have to prove that $g(x,y)=f(\sqrt{x^2+y^2})$ is differentiable.  I have to look for partial derivatives first :  $$\frac{\partial g}{\partial x}= f'(\sqrt{x^2+y^2}).\frac{2x}{2\sqrt{x^2+y^2}}=f'(\sqrt{x^2+y^2}).\frac{x}{\sqrt{x^2+y^2}}$$  $$\frac{\partial g}{\partial y}= f'(\sqrt{x^2+y^2}).\frac{2y}{2\sqrt{x^2+y^2}}=f'(\sqrt{x^2+y^2}).\frac{y}{\sqrt{x^2+y^2}}$$  I have used here that $f$ is differentiable on $\mathbb{R}$  As $f$ is continuously differentiable, we have $f'$ is continuous and I know that :  $\frac{x}{\sqrt{x^2+y^2}}$ and $\frac{y}{\sqrt{x^2+y^2}}$ are also continuous.     As $f'$ and $\frac{x}{\sqrt{x^2+y^2}}$ are continuous So is the product $\frac{\partial g}{\partial x}=f'(\sqrt{x^2+y^2}).\frac{x}{\sqrt{x^2+y^2}}$      As $f'$ and $\frac{y}{\sqrt{x^2+y^2}}$ are continuous So is the product $\frac{\partial g}{\partial y}=f'(\sqrt{x^2+y^2}).\frac{y}{\sqrt{x^2+y^2}}$   So we have continuous partial derivatives for $g(x,y)$..  So, I am willing to conclude that:     $g(x,y)$ is  differentiable on $\mathbb{R}^2$   I would be thankful if some one can confirm if the answer correct.  I would be much more thankful if some one can let me know if there are any gaps in my argument.  This is just a proof verification Question..  Thank you. I have two JSON objects with the same structure and I want to concat them together using Javascript. Is there an easy way to do this?