Problems on Pythagorean triangle - sides in arithmetic (geometric) progression 
Show that there is one (no) Pythagorean triangle whose sides are in arithmetic (geometric) progression.

The problem has two parts. There is one Pythagorean triangle whose sides are in arithmetic progression. There is no Pythagorean triangle whose sides are in geometric progression.
I know that
A triplet of positive integers x, y, z satisfying the equation $x^2+y^2=z^2$ is called Pythagorean triplet. If $x$, $y$, $z$ are Pythagorean triplet, then $x$, $y$, $z$ are pairwise prime to each other. 
I don't know whether Pythagorean triplet and Pythagorean triangle are same. 
Please help me to solve the problem. 
 A: If the sides are in arithmetic progression, WLOG we can choose the sides to be $a-d,a,a+d$  where $a,a-d>0\iff a>d$
So, $(a-d)^2+a^2=(a+d)^2\iff a^2=(a+d)^2-(a-d)^2=4ad$
$\iff a(a-4d)=0\implies a=4d$ as $a\ne0$
For Geometric Progression, let the sides be $a,ak,ak^2$ where $a>0,k\ge1$
So, $a^2+(ak)^2=(ak^2)^2\implies k^4-k^2-1=0$ as $a\ne0$
$\implies k^2=\dfrac{1\pm\sqrt5}2\implies k^2=\dfrac{\sqrt5-1}2$ as $k^2>1$
A: For the first part If $x,y,z$ are in arithmetic progression, then 
$$2y=x+z$$
or the same with $x,y$ interchanged.
Then 
$$x^2+y^2=z^2 \Rightarrow \\y^2=z^2-x^2=(z-x)(z+x)=(z-x)2y$$ 
This gives
$$x+z=2y \\
z-x=\frac{y}{2}$$
This gives $x=\frac{3y}{4}$ and $z=\frac{5y}{4}$. Deduce that $4|y$ and the writing $y=4k$ you get the arithmetic progression $(3k,4k,5k)$ is the general solution. 
For geometric, same ides:
$$ y^2=xz \Rightarrow x^2+xz-z^2=0$$
You can either solve the quadratic in $x/z$ or if you write $d=gcd(x,z)$ and write $x=dx', z=dz'$ with gcd $(x',z')=1$ you get 
$$x'^2+x'z'-z'^2=0$$
This implies $x'|z'^2$ and $z'|x'^2$ which combined with $gcd(x',z')=1$ yields $x'=z'=1$, which is not possible.
A: An approach from the other side.
Any Pythagorean triple can be generated from two non zero integers $m,n$ by:
$$
a=m^2-n^2 \qquad b=2mn \qquad c=m^2+n^2
$$ 
For an arithmetic progression we need:
$$
b-a=2mn-m^2+n^2=m^2+n^2-2mn=c-b 
$$
and solving we find $2n=m$ so that all such triple have the form 
$$
a=3n \qquad b=4n \qquad c=5n
$$
For a geometric progression we have:
$$
\dfrac{b}{a}=\dfrac{2mn}{m^2-n^2}=\dfrac{m^2+n^2}{2mn}=\dfrac{c}{b}
$$
that become $m^4-4m^2n^2-n^4=0$ and have no integer solutions since $\Delta'=n^4\sqrt{5}$
