So, we are trying to find all the solutions to $2^x + 3^y = z^2$ in nonnegative integers. Here are my insights:   First of all, $z^2$ can be either $0$ or $1$ modulo $3$. If $z^2 = 3k$, then LHS cannot be divisible by $3$ unless $y = 0$ and $x=2k+1$, and so we have $2^x+1 = z^2$, so one of the solutions is $(3,0,3)$ (but we still need to show that this is the only solution in this case).   Now let $y &gt; 0$, $x=2k$. In this case $2^x \equiv 1\ (\mod\ 3)$ and $z^2 \equiv 2^x (\mod 3)$. We have $$ 3^y = (z-2^k)(z+2^k), $$ so $(z-2^k)=3^a$ and $(z+2^k)=3^b$, so $3^b-3^a = 2^{k+1}$. If $a &gt; 0$, then $2^{k+1}$ must be divisible by $3$, but it is not, so suppose $a = 0$.  Now we have $$ 2^{k+1} = 3^b - 1, $$ Аnd this is where I stuck. It is easy to find two more solutions just iterating over $k$, but I do not know how to prove that there are no others. As far as I understand, we need to find some kind of an upper bound for $k$. Note: I am trying to avoid the use of Catalan's conjecture here.  Can you please help me to show that there are no other solutions to this equation and that $2^x+1=z^2$ has a unique solution?
Find all possible solutions of $$2^x + 3^y = z^2.$$    My approach.   First I substituted $x = 0$, and got the solution, then for $y = 0$. And for $x &gt;  0$ and $y &gt; 0$ , I just know the Pythagorean triple and got the solution as $(4,2,5)$ and $(4,2,-5)$. Please help me to solve it properly.
Basis for Skew Symmetric Matrix I'm trying to find a basis for the kernel for the following mapping:  Considering the linear transformation T: $M_{33} \rightarrow M_{33}  $ defined by $T(A) = .5(A + A^T)$. I know that this is basically asking for the basis under the condition that $T(A)=0$ which means that $A+A^T=0$ so $A^T = -A$.  I found that matrices that fit this condition are Skew Symmetric Matrices.  However, I'm not sure how to find the basis for the kernel of these matrices. Does a capacitor store charge? There're some easy concepts which I don't get exactly in my mind. I'm afraid I've been studying these things for two years of my engineering but they still bother me. Capacitor is one of them. Can someone explain?    What does a capacitor do? Does it store charges? If so, then how does it do?   I have searched it on Google and Yahoo but didn't find any helpful thing there (for me). So I will be glad if I got my problem solved here.  P.S. 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I need to enclosure a text block into big square brackets.  Something like this (I need to replicate it, more or less):   I've read that it could be possible with tikz but I don't know anything about that package.  Even, I found a code that reproduce something similar but with braces in this post:  The code is this:  \documentclass{article} \usepackage{lipsum} \usepackage[many]{tcolorbox} \usetikzlibrary{decorations.pathreplacing}  \newtcolorbox{rightbrace}{%     enhanced jigsaw,      breakable, % allow page breaks     frame hidden, % hide the default frame     overlay={%         \draw [             fill=none, % fill paper             decoration={brace,amplitude=0.5em},             decorate,             ultra thick,             gray,         ]         % right line         (frame.north east)--(frame.south east);     },     % paragraph skips obeyed within tcolorbox     parbox=false, }  \begin{document}  \begin{itemize}     \item First line           \begin{rightbrace}             \item Second line              \item Third line, which is quite long and seemingly tedious in the extreme             \item Fourth line, which isn't as long as the third            \end{rightbrace}     \item Fifth line \end{itemize}  \begin{rightbrace}     \lipsum[1] \end{rightbrace} \lipsum[2]  \end{document}   How could I put the square brackets around a block of text? 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I include the source, in the case that it is useful to anyone.  \begin{tikzpicture} \draw[help lines, gray, dotted] (-2,-2) grid (2,2);  \begin{scope}[shift={ (0,0) } ]     \begin{scope}[rotate=90]         \draw[thick] (-1,0) arc[start angle=-90, delta angle=90, %                               x radius=1, y radius=1] -- (0,1) %NW                      (0,-1) arc[start angle=180, delta angle=-90, %                                    x radius=1, y radius=1] -- (1,0); %SE         \draw[gray] (0,0) circle (1);     \end{scope} % Anchor replacements     \node(N1)       at (0,1)    {};     \node(S1)       at (0,-1)   {};     \node(E1)       at (1,0)    {};     \node(W1)       at (-1,0)   {}; % Label     \node(Crossing number) at (0,0) {1}; \end{scope}  \path[thick, gray, dotted] [ out=90,   in=270]     (N1.center) edge (0,2) [ out=180,  in=0]       (W1.center) edge (-2,0) [ out=270,  in=90]      (S1.center) edge (0,-2) [ out=360,  in=180]     (E1.center) edge (2,0); \end{tikzpicture}   PS. 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